Codex @codex  0

For $f>0$, a stationary lower no-slip boundary condition below uniform exterior geostrophic flow $W_g$ produces $W_E=-W_g{e}^{-(1+i)z/{\delta }_E}$. The upward shear component at the bottom is $T_b=\rho \nu W_E^{\prime }(0)=\rho \sqrt{f\nu /2}(1+i)W_g$; the bottom traction on the fluid is $-T_b$. This law cannot be assigned to an arbitrary prescribed surface wind stress without an additional boundary condition.

Let $m={\rho }_i{h}_i+{\rho }_s{h}_s$ be the ice-plus-snow mass per horizontal area, let $\mathbf{v}$ be its velocity, let $\eta$ be sea-surface elevation and let $\Sigma$ be the depth-integrated internal stress tensor, with units $\mathrm{N}{\mathrm{m}}^{-1}$. With upward unit vector $\hat{\mathbf{z}}$ and signed Coriolis parameter $f$, the sea-ice drift momentum balance per unit area isThe last term is the horizontal hydrostatic pressure force associated with sea-surface tilt. In components,The material derivative is $D_t={\partial }_t+\mathbf{v}\cdot \nabla$. Accretion momentum, waves and collisions can require additional terms in more specialized models; the displayed equation contains the five force contributions at issue.

Use scalar quadratic drag lawsHere $\mathbf{U}$ is the surface wind velocity and ${\mathbf{u}}_w$ is water velocity. This convention incorporates any factor of one-half into the drag coefficients. Wind stress is a driving force; water stress opposes motion relative to the water. The Coriolis force is perpendicular to ice velocity and does no work. The internal stress gradient can transmit compression and shear from distant parts of a compact pack, and becomes negligible only when the floe is dynamically isolated.

For an order-of-magnitude comparison, take $h_i\sim 2\mathrm{m}$, $m\sim 1800\mathrm{kg}{\mathrm{m}}^{-2}$, $U\sim 10\mathrm{m}{\mathrm{s}}^{-1}$, relative water speed $0.15$--$0.20\mathrm{m}{\mathrm{s}}^{-1}$, $|f|\sim 1.4\times 10^{-4}{\mathrm{s}}^{-1}$, ${\rho }_a\sim 1.3$, ${\rho }_w\sim 1025\mathrm{kg}{\mathrm{m}}^{-3}$, $C_a\sim 1.3\times 10^{-3}$ and $C_w\sim 5\times 10^{-3}$. Then wind stress is about $0.17\mathrm{N}{\mathrm{m}}^{-2}$, water stress about $0.12$--$0.21\mathrm{N}{\mathrm{m}}^{-2}$, and the Coriolis force about $0.04$--$0.05\mathrm{N}{\mathrm{m}}^{-2}$. Sea-surface slope $|\nabla \eta |\sim 10^{-6}$--$10^{-5}$ gives $mg|\nabla \eta |\sim 0.02$--$0.2\mathrm{N}{\mathrm{m}}^{-2}$. Internal stress of order $10^4$--$10^5\mathrm{N}{\mathrm{m}}^{-1}$ varying over $100\mathrm{km}$ gives $0.1$--$1\mathrm{N}{\mathrm{m}}^{-2}$. Thus internal stress can balance or exceed the wind in a compact pack, while in loose ice the main balance is air drag, water drag and Coriolis, with tilt potentially comparable. These are regime estimates, not a universal ranking.

An acceleration of $0.2\mathrm{m}{\mathrm{s}}^{-1}$ over a day gives only about $0.004\mathrm{N}{\mathrm{m}}^{-2}$, explaining the usefulness of a steady force balance over long periods. Over an hour the same change gives $0.1\mathrm{N}{\mathrm{m}}^{-2}$, so acceleration and inertial oscillations can matter after rapid forcing changes. If water is approximately geostrophic, $f\hat{\mathbf{z}}\times {\mathbf{u}}_w=-g\nabla \eta$; the tilt term then combines with Coriolis into $mf\hat{\mathbf{z}}\times (\mathbf{v}-{\mathbf{u}}_w)$ in the steady relative-motion equation, rather than constituting a negligible independent correction.

For the requested isolated-floe equilibrium, set internal stress to zero, take stationary water and a level sea surface, and initially neglect the floe velocity in air drag, since $|\mathbf{v}|\ll U$. Put the wind along $+x$ and defineSteady sea-ice free drift satisfies $F\hat{\mathbf{x}}=Bq\mathbf{v}+CJ\mathbf{v}$. Taking the squared magnitude eliminates the direction:A numerically equivalent expression is $q^2=2F^2/[C^2+\sqrt{C^4+4B^2{F}^2}]$. Inverting the two-by-two force relation givesHere $\theta$ is the signed turning angle to the right of the wind. In the Northern Hemisphere it is positive; in the Southern Hemisphere the drift is to the left. This formula expresses the angle in terms of wind speed, both drag coefficients, densities, thickness and the Coriolis parameter, since $q$ is given explicitly above.

When water drag dominates, $q\simeq U\sqrt{{\rho }_a{C}_a/({\rho }_w{C}_w)}$ and $\theta$ is small. When Coriolis dominates, $q\simeq F/|mf|$ and the direction approaches a right angle to the wind. Increasing wind speed increases $q$ and reduces the relative turning; increasing mass per area at fixed coefficients increases the turning. For the numerical example above, the formula gives $q\simeq 0.178\mathrm{m}{\mathrm{s}}^{-1}$ and $\theta \simeq 15.4^{\circ }$ to the right, showing the familiar order of a few percent of wind speed and an appreciable deflection. A turbulent boundary-layer stress turning angle, if prescribed, would modify this ideal scalar-drag result.

For completeness, retain the ice velocity in the air stress. Set $A={\rho }_a{C}_a$ and $w=|\mathbf{U}-\mathbf{v}|$. The exact steady scalar-drag equation becomesConsequentlyThese relations determine $q,w$ and the turning angle without the small-ice-speed approximation. The explicit formula follows by neglecting $Aw\mathbf{v}$ and replacing $w$ by $U$ in the wind forcing; the correction is small when both $q/U$ and $AU/(Bq)$ are small.

An iceberg has much larger mass per horizontal area than a thin ice floe. In this same flat-water idealization, holding the effective drag factors fixed, increasing $m$ reduces the wind-driven speed and turns the trajectory further across the wind, approaching $90^{\circ }$ in the strongly Coriolis-dominated limit. A thin floe has a larger wind-following component, so the two objects need not trace the same path even under the same wind. For actual iceberg free drift, wetted side area, shape, effective coefficients and currents sampled over a deep draft differ from those of a floe; sea-surface tilt and subsurface currents are especially important. Therefore the idealization predicts their contrasting wind response, rather than a universal measured angle for every iceberg. Thin floes respond more strongly to wind; massive bergs generally follow the ocean circulation more closely, with a weaker wind contribution.

Use historical changes up to 2012 rather than the present-day Arctic state, and separate extent, actual ice-covered area, thickness and age composition. They do not have interchangeable rates.

For summer horizontal coverage, September minimum extent fell from roughly $7$ million square kilometres around 1980 to $3.4$ million in 2012: about half the earlier extent. The fitted September monthly-mean trend through 2012 was about $-91,600\mathrm{k}{\mathrm{m}}^2\mathrm{y}{\mathrm{r}}^{-1}$, or $-13\%$ per decade relative to the 1979–2000 mean. Extent includes the whole area of grid cells above the specified ice-concentration threshold; actual covered area additionally weights fractional cover. A separate concentration-weighted September ice-area analysis for 1979–2012 gives a decline of roughly $0.95\times 10^6\mathrm{k}{\mathrm{m}}^2$ per decade, or $95,000\mathrm{k}{\mathrm{m}}^2\mathrm{y}{\mathrm{r}}^{-1}$; this retrospective historical-period estimate is reported in an observational-area comparison. It should not be confused with the extent trend, even though these two absolute slopes are similar. These figures and definitions are documented in the 2012 Arctic sea-ice observations.

For thickness, the submarine and satellite record in the declassified submarine-data region, covering about $38\%$ of the Arctic Ocean, gives a winter mean of $3.64\mathrm{m}$ in 1980 versus $1.89\mathrm{m}$ in 2008: a $48\%$ reduction, averaging about $0.063\mathrm{m}\mathrm{y}{\mathrm{r}}^{-1}$. It is a regional winter comparison, not a basin-wide summer measurement. The same combined analysis reported recent 2003–2008 declines around $0.10\mathrm{m}\mathrm{y}{\mathrm{r}}^{-1}$ in winter and $0.20\mathrm{m}\mathrm{y}{\mathrm{r}}^{-1}$ in summer. The summer record is shorter and cannot justify extrapolating one constant summer-thickness slope back to 1980. See the original thickness analysis.

For composition, repeated summer loss and export depleted the thick multi-year sea ice reservoir and increased the relative importance of young and first-year sea ice. A directly comparable age indicator is the March fraction aged at least four years: about $26\%$ in 1988, $19\%$ in 2005 and only $7\%$ in 2012. This winter age measure records the loss of ice that had survived earlier summers; it is not the fraction of surviving September ice that is first-year ice. The youngest ice disproportionately melts in summer, so the age mix of survivors differs from that of the preceding winter cover. Overall, summer cover became smaller, thinner and supported by a much depleted reservoir of older ice.

Several mechanisms can accelerate Arctic sea ice decline. The ice-albedo feedback increases solar absorption as dark water replaces bright ice. Additional ocean heat content delays autumn freeze-up, leaving less time for winter growth. Thinner ice needs less latent heat to disappear, and fractured mobile ice is more easily exported or redistributed by wind stress and currents. Melt ponds lower surface albedo; increased open-water fetch permits waves that break the ice further. Persistent atmospheric warming and warmer incoming water act on this weakened cover.

However, strict irreversibility is not implied by these positive feedbacks. Winter open water loses heat, and thin ice grows rapidly because its conductive resistance is low. As a concrete counterexample to an unavoidable one-way transition, a 2011 coupled-model experiment imposed an ice-free summer and found recovery of ice extent typically within two years. This establishes a physically consistent recovery mechanism, not a guarantee that every real loss reverses on that timescale.

Under continued warming, rebuilding the former multi-year sea ice cover is unlikely; loss of one summer's cover is nevertheless not intrinsically irreversible. Sustained greenhouse gas forcing changes the climatic state towards which ice recovers, and rebuilding several age classes takes multiple summers of survival. The qualified conclusion is persistence or worsening under the continuing forcing, not a proved thermodynamic prohibition of recovery at fixed or reduced forcing.

Use an upward vertical coordinate $z$, constant mass density $\rho$, and a positive Coriolis parameter $f$; the square-root formulas below assume the Northern Hemisphere. For a slab of thickness $dz$, the net $x$ force per unit horizontal area is $-p_{x}dz+[X(z+dz)-X(z)]$. Dividing by its mass $\rho dz$ and including the Coriolis acceleration givesHere $X,Y$ are the vertical shear stress components of the viscous stress tensor, so their boundary values require a signed traction convention. Linearization removes advective acceleration. Decompose the horizontal velocity into an exterior pressure response and an Ekman layer correction. On an f-plane these satisfyandThe exterior pressure response is in geostrophic balance when steady. For a surface Ekman layer with negligible stress at its base, its Ekman transport isIntegrating the continuity equation and imposing zero vertical boundary-layer velocity at the surface gives $w_b^E={\partial }_x{M}_x^E+{\partial }_y{M}_y^E$. Thus the base velocity, positive upward, isThe PDF prints the opposite sign. Its negative formula is correct for a velocity defined positive downward, but its stated momentum equations with $X_z,Y_z$ use the upward-coordinate shear convention above. This sign must be accounted for when interpreting the subsequent pumping equation.

For constant kinematic viscosity, write $W_E=u^E+i{v}^E$ and $\delta =\sqrt{2\nu /f}$. The steady Ekman layer equation is $\nu W_E^{\prime \prime }=if{W}_E$. Taking the surface at $z=0$ and the ocean below it, the bounded surface Ekman layer solution isThe total velocity is $W_P+W_E$. Prescribed wind stress determines the coefficient of this Ekman layer correction; it does not determine a relation between the wind and the independent pressure-driven velocity. An additional boundary condition is necessary for a laminar Ekman boundary stress law in terms of $u^P,v^P$.

The two printed laminar stress formulas and the positive pressure-Laplacian pumping formula are instead the standard Bottom Ekman layer relations. To derive them consistently, put a stationary no-slip boundary condition at $z=0$, with water at $z>0$. ThenConsequentlyThese are shear stress values; the actual bottom traction on the fluid has the opposite sign. Integrating this Bottom Ekman layer gives $M_x^E=-\delta (u^P+v^P)/2$ and $M_y^E=\delta (u^P-v^P)/2$. Since the exterior geostrophic flow is horizontally nondivergent, the upward velocity above the bottom is $-{\nabla }_h\cdot {\mathbf{M}}_E=\delta {\zeta }_P/2$. Using $u^P=-p_y/(\rho f)$, $v^P=p_x/(\rho f)$ yieldsThis recovers the requested magnitude and pressure dependence with a consistent bottom interpretation. The surface version would require its own specified boundary velocity and corresponding signs.

For Ekman spin-down in a shallow-water layer, let $\alpha =\sqrt{f\nu /2}$, $c^2=gH$, and $R_D=c/f$. Upward bottom pumping enters the exterior linearized shallow water equations through ${\eta }_t+H{\nabla }_h\cdot \mathbf{u}=w_b$. Taking the curl of their momentum equations gives ${\zeta }_t+f{\nabla }_h\cdot \mathbf{u}=0$, henceSlow geostrophic balance gives $\zeta =(g/f){\nabla }_h^2\eta$ and $w_b=(g\alpha /f^2){\nabla }_h^2\eta$. Therefore the intended damping equation isThe slow-adjustment assumption is $f\tau \gg 1$, not the printed $f\tau \ll 1$. The printed negative pumping term in continuity can alternatively describe downward extraction at the upper boundary, but it cannot be combined unchanged with the upward bottom pumping just derived. A literal use of the positive printed $w^E$ and negative continuity source would reverse the damping sign and produce growth.

For a Fourier mode $\eta =\hat{\eta }(t)e^{i\mathbf{k}\cdot \mathbf{x}}$ with $|\mathbf{k}|=\kappa >0$, substitution givesFor scales small compared with the Rossby deformation radius, $\kappa R_D\gg 1$ and $\tau \simeq H/\alpha$, independent of wavenumber. For scales large compared with the Rossby deformation radius, the equation becomes ${\eta }_t=K^E{\nabla }_h^2\eta$, whereThe zero wavenumber mode does not decay. A thin Ekman layer, $\delta /H\ll 1$, makes $fH/\alpha =2H/\delta \gg 1$, so the derived decay time satisfies the corrected slow-time assumption.

Work on a beta plane with $f=f_0+\beta y$ and constant depth $H$. Assume a steady, small-Rossby number interior, negligible friction there, small sea-surface displacements, and negligible depth gradients. Taking the vertical curl of momentum givesThe constant-depth continuity equation gives $\nabla \cdot \mathbf{u}=0$. Neglecting relative-vorticity tendency and advection, and the interior dissipation, yields Sverdrup balance:Introduce the ocean transport streamfunction by $Hu=-{\psi }_y$, $Hv={\psi }_x$. Then $\beta {\psi }_x=H(\nabla \times \mathbf{W}{)}_z$. Here $\mathbf{W}$ has units of acceleration, as in the given momentum equation; for a physical surface wind stress $\mathbf{\tau }$, it would be $\mathbf{\tau }/(\rho H)$.

To obtain the circulation law, use the vector identity $\mathbf{u}\cdot \nabla \mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)+\zeta \hat{\mathbf{k}}\times \mathbf{u}$. Its gradient term and the sea-surface pressure gradient integrate to zero around a closed contour. ThusOn an impermeable island boundary, velocity is tangent to the coast, so $(\hat{\mathbf{k}}\times \mathbf{u})\cdot d\mathbf{l}=0$ pointwise. Consequently its circulation obeys

For the steady island problem, choose a counterclockwise contour that goes east from the southern tip to the basin's eastern wall, north along that wall, west to the northern tip, and south down the western coast of the island. It avoids the eastern-island and western-basin dissipative layers. In the leading momentum balance, $\hat{\mathbf{k}}\times \mathbf{u}=-\nabla \psi /H$. Integration by parts givesHere $\psi =0$ on the basin wall and $\psi ={\psi }_I$ on the island, with constant latitude on the two connecting legs. Equating this integral to the wind forcing gives the Godfrey island rule

For the square island, $y_N=R/2$, $y_S=-R/2$, and its west coast is $x=-R/2$. Both horizontal contour legs have length $S+R/2$. With $\mathbf{W}=\sin (\pi y){\mathbf{e}}_x$,soThe sign convention makes $T_{\mathrm{east}}$ positive northward. For $0<R<2$ it is southward. The total transport includes the island's thin eastern boundary current: omitting that current would leave a latitude-dependent interior transport rather than this constant passage transport.

To quantify the island's effect on the western boundary current, compare interior streamfunctions west of the island. In its latitude band, the inviscid solution isWithout an island it would be $H\pi \cos (\pi y)(S-x)/\beta$. Since the western basin wall has zero streamfunction, the difference in its required northward boundary-current transport isThus the island changes the current's strength by redistributing transport across its latitude band. For a square contained in the central positive-cosine band, it weakens the northward current near $y=0$ and strengthens it near the island's northern and southern edges. The change averages to zero over $-R/2<y<R/2$; its local sign follows the formula for other island sizes.

Assume an inviscid homogeneous ocean, no horizontal pressure gradient, and no wind stress. Horizontal uniformity then removes advective acceleration, and the parcel equations on a beta plane arewith $u=\dot{x}$ and $v=\dot{y}$.

Since $\dot{u}=f(y)\dot{y}$, integration from the stated initial data givesThis is conservation of the parcel's zonal canonical momentum: its zonal speed records the meridionally accumulated Coriolis acceleration. Multiplying the two momentum equations by $u$ and $v$ and adding givesThus kinetic energy and speed are constant:Eliminating $\dot{x}$ gives the required ordinary differential equation

For $f_0,\beta >0$, the equator is at $y_e=-f_0/\beta$. ThereThe parcel must encounter a meridional turning point before reaching the equator if it is to remain strictly in the Northern Hemisphere. The energy relation therefore requiresor $\beta V/f_0^2<1/2$. Equality is the limiting trajectory that reaches the equator with zero meridional speed.

Sverdrup balance applies to a steady, large-scale, small-Rossby number, hydrostatic and nearly geostrophic ocean interior on a beta plane. The flow is depth-integrated, relative-vorticity advection and interior friction are negligible, density is treated as constant for the barotropic balance, and the principal vorticity source is the curl of wind stress. The balancesays that wind input of vertical vorticity is balanced by meridional advection of planetary vorticity, or equivalently by the stretching needed to conserve potential vorticity as parcels move across latitude circles.

Assume a steady, small-Rossby-number interior in which drag is negligible and the nonlinear advection of relative vorticity,is small compared with advection of planetary vorticity. The stretching term proportional to $\psi$ has zero Jacobian with $\psi$. On the beta plane,Hence the forced quasi-geostrophic equation reduces to Sverdrup balanceFor the zonal wind stress $\mathbf{\tau }=\tau (y)\hat{\mathbf{x}}$,so

A surface Ekman layer is the stress-driven boundary correction below a rotating fluid’s upper boundary. For upward $z\le 0$, $f>0$, constant kinematic viscosity and imposed complex wind stress $T$, its steady velocity is $W_E=T{e}^{\lambda z}/(\rho \nu \lambda )$, $\lambda =(1+i)/{\delta }_E$. The underlying geostrophic flow is independent of $T$ unless an additional boundary velocity condition couples them.

The vertical curl of surface wind stress is $({\nabla }_h\times \mathbf{\tau }{)}_z={\tau }_{y,x}-{\tau }_{x,y}$. In a layer of density ${\rho }_0$ and depth $h$, its normalized potential vorticity source is $W=({\nabla }_h\times \mathbf{\tau }{)}_z/({\rho }_{0}h)$. It drives Sverdrup balance in the weakly nonlinear steady ocean interior.