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Take $x$ positive in the off-ice wind direction, from the compact pack towards the open sea. The wind initially separates the outer ice floes, creating irregular polynyas. A larger opening gives more open-water fetch, so stronger short wind waves develop before reaching its downwind edge. Their reflection supplies a positive force on the downwind ice floes; these catch their neighbours and compact into an ice-edge band. Incoming longer swell exerts force in the opposite direction. The short waves can exert substantial force despite their smaller amplitude because a small ice floe reflects them much more effectively than it reflects the long swell.

The surface-gravity-wave energy of the short waves grows with fetch in the windward polynya, then decays rapidly across the band. Swell enters from the seaward side and usually attenuates more slowly. This gives inward forcing from the two sides: short-wave forcing is largest at the windward face and swell forcing is largest at the seaward face. In the next open polynya, short wind waves regrow from the weak transmitted component. Reflected waves also enhance energy locally on the incident side, with interference that is omitted from a smooth, phase-averaged sketch.

Each plotted component is normalized by its own incident energy; the curves do not assert equal absolute wind-wave and swell energies. The band is partially transmitting. For a perfectly opaque reflector, the transmitted component would instead vanish.

The initial bands have unequal floe inventories, widths and forcing. Differential drift and collisions merge some into composite bands. Larger bands tend to shield smaller downstream accumulations from the short-wave forcing needed to keep them separate, while sufficiently wide intervening polynyas can generate fresh wind waves and maintain separation. Finite available ice, available wave fetch, the opposing swell, wind strength and duration, floe size and thickness, and subsequent mergers limit the number of persistent bands. The wave-force formula alone does not select a universal count. This mechanism and its merger interpretation are supported by the original ice-band study.

There is a normalization issue in the requested stress calculation. Let $\mathcal{E}={\rho }_{w}g{a}^2/2$ be standard linear surface-gravity-wave energy for crest amplitude $a$. The deep-water wave radiation stress is $S_{xx}=\mathcal{E}/2$. Hence standard momentum balance givesFor lossless reflection with amplitude reflection coefficient $R$, $r^2=R^2{a}^2$ and $t^2=(1-R^2)a^2$, so $F_{\mathrm{phys}}={\rho }_{w}g{R}^2{a}^2/2$.

In the usual independent-floe, weak-reflection approximation, there are about $p/d$ effective layers per unit distance. Each removes a fraction $R^2$ of the forward energy, givingThis yields the intended wave-driven ice-band compaction:The attenuation approximation retains the leading term in $R^2$; an independent discrete-layer model instead gives an energy coefficient $-(p/d)\log (1-R^2)$. Neither coefficient follows from the single-object force formula without this additional scattering closure.

For ordinary crest amplitudes, the force printed in the paper is $4F_{\mathrm{phys}}$. With that printed normalization and the same attenuation law, its derivative is four times the requested stress. More generally, if $a^2=a_0^2{e}^{-\kappa x}$, the printed lossless force gives $-F^{\prime }=2{\rho }_{w}g{R}^2\kappa a^2$. Recovering the stated stress from it requires $\kappa =p{R}^2/(4d)$ instead, a different unspecified attenuation convention. The supplied force and stress cannot both be derived from standard amplitudes and the usual floe-layer attenuation law. Restoring the factor $1/4$ in the force gives a consistent intended model.

Using the printed force for the numerical question, perfect reflection gives $r=a$, $t=0$, and thereforeUsing the requested stress formula for partial reflection givesFor comparison, the consistently normalized perfect-reflection force is $50.3\mathrm{N}{\mathrm{m}}^{-1}$, and the printed force with standard attenuation would give $0.628\mathrm{N}{\mathrm{m}}^{-2}$ for the partial-reflection derivative. The approximate stress is a force per horizontal area, not a direct measure of the three-dimensional ice-skeleton stress.

The inward force gradient helps maintain a coherent, close-packed band, especially near the incident-wave faces. It is modest enough that wind, current, swell changes or mergers can disrupt the arrangement; it does not guarantee permanent mechanical stability. Perfect reflection estimates a bounding force, whereas the much smaller partial-reflection stress varies as $R^4$. No mechanical-strength law is supplied, so stability can be assessed only qualitatively.

Take $z$ positive upwards and let $f$ denote axial body force per unit volume. With inertia and surface tension omitted, the extensional equations for a slender Newtonian column areMass conservation gives the first equation. The second uses the Trouton ratio three: the leading radial velocity is $u_r=-r{w}_z/2$, the radial normal traction gives $p=p_{\mathrm{ext}}-\mu w_z$, and the axial stress is $-p_{\mathrm{ext}}+3\mu w_z$. An axial control volume balance then gives the displayed equation. The approximation requires a slender column and slowly varying radius.

Let $h=R-a$. Since the container base is closed, the total volume flux at every section is zero. The column carries flux $\pi a^{2}w$, so the annulus carries $-\pi a^{2}w$ and has mean speed of order $a|w|/h$. Its interfacial shear stress is therefore of order $\lambda \mu a|w|/h^2$. Transmitting this stress through the core produces an axial velocity variation of order $\lambda a^2|w|/h^2$. The plug-flow criterion for a column in a low-viscosity annulus is consequently

Use modified pressure $P=p_{\mathrm{ext}}+\rho gz$, removing the annular fluid's hydrostatic pressure. In a local planar gap, with $y=0$ at the column and $y=h$ at the container wall, lubrication theory givesThe required return flux per unit circumference is $q_g=-aw/2$ to leading order in $h/a$. Its pressure-driven component is larger than the Couette component by $a/h$, so the annular return flow around a slumping column givesThe annular shear contribution to the integrated axial force is smaller than this pressure-gradient term by $h/a$. Retaining the leading terms and substituting $f=-(\rho +\Delta \rho )g$ gives the dimensional equationsThe boundary conditions are $w(0,t)=0$ at the base and $w_z(L(t),t)=0$ at the free upper end; the latter expresses zero excess axial extensional stress.

For $\lambda =0$ and initially constant $a=a_0$, put $G=\Delta \rho g/(3\mu )$. Then $w_{zz}=G$, so

For $\lambda >0$, defineThis function records the leading thin-annulus resistance; differences from curvature and the Couette term enter at the already neglected relative order $h/a$. Choose the extensional screening length of a slumping column and the associated scales asHere $h_0=R-a_0$ denotes the initial annular gap in this question. With $Z=z/\hat{z}$, $T=t/\hat{t}$ and $W=w/\hat{w}$, the dimensional equations reduce to

Initially $\varphi ={\varphi }_0$ and ${\Lambda }_0=L_0/\hat{z}$, so $W_{ZZ}=1+W$, with $W(0)=0$ and $W_Z({\Lambda }_0)=0$. The initial velocity profile of an annularly confined column isThe column thickens everywhere below its stress-free top, while that initial thickening rate is zero at the top.

For a tall column, ${\Lambda }_0\gg 1$, the limiting forms away from exponentially small end corrections areThe bulk falls at approximately $-\hat{w}$, where excess weight balances annular hydraulic resistance. Extensional stresses adjust this velocity to zero in a basal boundary layer of thickness $\hat{z}$; almost all thickening occurs there.

For a short column on this scale, ${\Lambda }_0\ll 1$, while still slender, the limits areThe annular-resistance term $W$ is small, so excess weight is balanced predominantly by extensional viscous stress. The top speed is $-\hat{w}{\Lambda }_0^2/2$, recovering the zero-annular-viscosity result. Thus $\hat{z}$ is the vertical distance over which extensional stress communicates the basal constraint before annular resistance screens it. The sketches below use separate natural normalizations for the two limits.

At leading order in the lubrication approximation, surface arclength is horizontal distance. An insoluble surfactant is transported by the surface velocity, without bulk exchange or diffusion. Conservation of insoluble surfactant on a moving interface therefore becomesThe local hydrostatic pressure and capillary normal stress giveSolve $\mu u_{zz}=p_x$ with no-slip boundary condition at $z=0$ and Marangoni stress $\mu u_z(h)={\gamma }_x=-A{C}_x$. The velocity, surface velocity and volume flux per unit width areMass conservation, $h_t+q_x=0$, yields the thin-film equations with insoluble surfactant:and the corresponding surfactant equation isThe hydrostatic and capillary terms thus affect both the film flux and the surface transport, with different coefficients.

For finite-mass Marangoni spreading on a liquid film, first neglect both pressure-gradient terms. If the spread has size $\ell$, then $C\sim M/\ell$ and $C_x\sim M/{\ell }^2$, giving surface speed $u_s\sim AM{h}_0/(\mu {\ell }^2)$. Equating $\ell /t$ to this speed givesSet $\ell =(AM{h}_{0}t/\mu {)}^{1/3}$ exactly as a similarity scale, and writeOn the positive half of the pool, substitution produces two ordinary differential equations:Symmetry fixes the flux integration constant in the second equation to zero, giving $H{\Gamma }^{\prime }=-\eta /3$ wherever $\Gamma >0$. Substitution into the first equation gives $\eta H^{\prime }=H$. Thus the linear similarity profiles for surfactant spreading have the formThe concentration vanishes at the moving front. The two integral constraints are total surfactant mass and conservation of the fluid volume relative to the undisturbed layer:The first gives $k={\eta }_N^2/3$; the second gives $k{\eta }_N=2$. Therefore ${\eta }_N^3=6$. In physical variables, the solution isOutside the pool the leading outer solution has $h=h_0$, $C=0$. In particular, $h(x_{N-},t)=2h_0$: the subscript means the one-sided limit just behind the front. As a consistency check, the surface speed there is $x_N/(3t)={\dot{x}}_N$ and the fluid jump satisfies the moving-front mass conservation condition. This ideal outer solution has a height jump from $2h_0$ to $h_0$.

That jump cannot persist in a physical interface with nonzero gravity or surface tension. Across a smoothing region of width $\Delta$, $h$ changes by $O(h_0)$. The Marangoni flux scales as $AM{h}_0^2/(\mu x_N^2)$. At $g=0$, its balance with the capillary flux ${\gamma }_0{h}_0^4/(\mu {\Delta }^3)$ gives capillary smoothing of a Marangoni front:Here the surface tension approaches ${\gamma }_0$ at the surfactant-free edge. The assumed concentration-gradient scale is $M/x_N^2$ even inside the smoothing region.

When gravity dominates capillarity, the hydrostatic flux is $\rho g{h}_0^4/(\mu \Delta )$. The same balance gives gravity smoothing of a Marangoni front:The gravity-dominated condition is ${\Delta }^2\gg {\gamma }_0/(\rho g)$, the square of the capillary length. The printed condition omits $\rho$; the expression above restores dimensional consistency.

This width grows faster than $x_N$. It becomes comparable with the pool size whenThe time is a scaling estimate, so numerical factors cannot be fixed by the width balance. For $t\gg t^{\ast }$, the sharp-front similarity profile no longer describes the film: hydrostatic pressure levels the layer over the whole pool, and its thickness tends towards $h_0$ with an increasingly small depression beneath the surfactant. The surface can still spread by Marangoni stress while a hydrostatic-pressure-driven return flow nearly cancels the net film flux. In this gravity-levelled surfactant film, $q\simeq 0$ gives $h_x\simeq -3A{C}_x/(2\rho g{h}_0)$, so the fractional height variation is of order $AM/(\rho g{h}_0^2{x}_N)\ll 1$.

Linearize about a motionless, isotropic, relaxed state by taking velocity, deformation and extra stress to be first order in a small amplitude. The advection $u\cdot \nabla \sigma$ and rotational commutator $\omega \sigma -\sigma \omega$ are then second order. The corotational Maxwell fluid therefore reduces to the linear Maxwell fluid:where $e$ is the rate-of-strain tensor. This is the linear Maxwell viscoelasticity model, mechanically a spring and dashpot in series.

For one shear component, dividing by $\mu$ gives $\dot{\gamma }=\sigma /\mu +(\lambda /\mu )\dot{\sigma }$. Comparing with a dashpot of viscosity $\mu$ in series with a spring of shear modulus $G$ identifiesIn particular a held strain with zero subsequent shear rate has $\sigma (t)=\sigma (0)e^{-t/\lambda }$. At long times under steady small shear, $\sigma =\mu \dot{\gamma }$, whereas a sufficiently rapid deformation probes the elastic modulus $G$.

In linear simple shear, $\dot{\gamma }$ has only off-diagonal $xy$ components. The diagonal stresses have no forcing and, from a relaxed initial state, stay zero. Any pre-existing diagonal stress decays homogeneously. Thus both normal-stress differences vanish in the linear response to shear. The finite-shear normal-stress differences generated by products of shear rate and stress are quadratic in deformation amplitude and have been discarded by this linearization. This conclusion is about simple shear, not about every possible extensional deformation.

Work in an inertial frame centred on the centre of mass, and follow a fixed material mass, with no stellar wind or accretion through its boundary. Assume nonrelativistic motion, Newtonian self-gravity, no external gravitational field, and sufficiently regular integrable fields. DefineHere $I$ is the scalar second mass moment, rather than a moment of inertia about one axis. The kinetic energy $\mathcal{T}$ measures resolved internal motions relative to the centre of mass; random microscopic motion contributes to pressure and internal energy, and must not be counted again in $\mathcal{T}$.

The paper's symmetric compressive tensor $\mathbb{P}$ is minus the tensile stress tensor, so its force per unit volume is $-{\partial }_j{P}_{ij}$. Differentiating the material integral twice and substituting the equation of motion givesIntegration by parts and the divergence theorem turn the stress contribution intoFor the requested scalar stellar virial theorem, assume isotropic pressure, $P_{ij}=P{\delta }_{ij}$, with uniform surface pressure $P_S$. Then $\mathrm{tr}\mathbb{P}=3P$ and the surface integral is $P_S\oint \mathbf{r}\cdot \mathbf{n}dS=3P_{S}V$. No spherical shape is needed for this step.

Finally, symmetrizing the pair integral for the Newtonian gravitational potential givesThereforeAnisotropic magnetic or viscous stresses retain their full trace and boundary terms; a changing integration mass adds transport terms.

The material derivative follows the interfacial surfactant. The term $-C{\nabla }_s\cdot {\mathbf{u}}_s$ describes dilution by tangential expansion and concentration by compression. The term $-C(\mathbf{u}\cdot \mathbf{n}){\nabla }_s\cdot \mathbf{n}$ accounts for changing surface area due to normal motion of a curved interface. Finally, $D_s{\nabla }_s^{2}C$ describes surface diffusion. Together these give conservation of insoluble surfactant on a moving interface.

The relevant surface Péclet number isWrite $S=\mathbf{n}\cdot \mathbf{A}\cdot \mathbf{n}$ and ${\mathbf{u}}_s=\mathbf{Ax}-S\mathbf{x}$. Since $\mathbf{A}$ has zero trace and ${\nabla }_s\mathbf{n}={\mathbf{I}}_s/a$, the surface divergence isThere is no normal motion. To first order in ${\mathrm{Pe}}_s$, the steady surfactant balance becomes $D_s{\nabla }_s^2{C}^{\prime }=C_0{\nabla }_s\cdot {\mathbf{u}}_s=-3C_{0}S$. The traceless quadratic $S$ is a degree-two spherical harmonic, so ${\nabla }_s^{2}S=-6S/a^2$. The mean of $C^{\prime }$ is zero by total surfactant conservation; consequentlyThe discarded advective term ${\mathbf{u}}_s\cdot {\nabla }_s{C}^{\prime }$ is smaller by ${\mathrm{Pe}}_s$. This is the quadrupolar surfactant distribution on a spherical interface.

With the unit normal directed from the bubble into the exterior, the interfacial stress balance with variable surface tension isUsing the specified first-order curvature and $\gamma ={\gamma }_0-K{\gamma }_{1}S$, together with ${\nabla }_{s}S=2{\mathbf{I}}_s\mathbf{An}/a$, givesThe first term is the spherical capillary pressure; the last contains the normal tension correction and the tangential Marangoni stress.

The ambient Stokes flow has the symmetry of the symmetric traceless tensor $\mathbf{E}$. In the Unscaled Papkovich–Neuber representation, ${\chi }_{\infty }=\mathbf{x}\cdot \mathbf{E}\cdot \mathbf{x}/2$ is harmonic and yields $\mathbf{Ex}$. The decaying vector potential must have the form $\mathbf{E}\nabla (1/r)$, and the scalar disturbance must have the degree-two form $\mathbf{E}:\nabla \nabla (1/r)$. They are harmonic outside the bubble and have precisely the required rotational covariance. Dimensionless amplitudes may therefore be writtenFor a steady bubble, the no-penetration boundary condition gives $1+P-3Q=0$. Its tangential velocity gives $\alpha =1+2Q$. The tangential stress boundary condition then yieldsHence $\alpha =5/(5+2M)$, $Q=-M/(5+2M)$ and $P=-(5+5M)/(5+2M)$. The inviscid interior supplies only the constant pressure balancing $2{\gamma }_0/a$. Matching the remaining normal stress givesEliminating $P,Q,\alpha$ givesAs $M\to \infty$, $\alpha \to 0$: the Marangoni stress suppresses tangential motion and effectively immobilizes the interface. This limit must retain the small surface Péclet number and small-deformation assumptions; $M$ can grow through increasing ${\gamma }_1$ without invalidating the linear concentration approximation.

Each broad face supplies surface tension $\gamma$ pulling the rounded hole edge into the remaining sheet. Their resultant is $2\gamma$ per unit circumference. At the inner boundary the fluid's outward normal is $-{\mathbf{e}}_r$, so the boundary traction is outward in the radial direction when $h{\sigma }_{rr}=-2\gamma$. The right panel of the preceding diagram shows these two capillary pulls.

With uniform thickness and $p_{\mathrm{ext}}=0$, the axisymmetric viscous-sheet stretching equations reduce toThis Euler-Cauchy equation gives $u=Ar+B/r$. The fixed outer rim imposes $u(R_0)=0$, hence $B=-A{R}_0^2$. Because $(ru{)}_r/r=2A$, conservation of mass gives $\dot{h}=-2Ah$, independent of $r$. Thus a uniform sheet remains uniform.

The radial stress is ${\sigma }_{rr}=6\mu A-2\mu B/r^2$. Its value at the hole edge determinesThe edge is material, so $\dot{R}=u(R)$ andNeglecting the initially tiny hole's volume, conservation of mass gives $\pi h(R_0^2-R^2)=\pi h_0{R}_0^2$, or $h=h_0/(1-x^2)$ with $x=R/R_0$. Therefore the capillary growth of a hole in a viscous sheet obeysFor a finite initial hole $x_i$, replace $h_0$ in the denominator by $h_0(1-x_i^2)$. A nonzero seed is needed: the exact initial condition $x(0)=0$ gives the stationary solution of this differential equation. For a small positive seed, $x$ initially grows exponentially at rate $\gamma /(\mu h_0)$.

For the stated axisymmetric flow, the diagonal components of the rate-of-strain tensor areTheir sum is $u_r+u/r+w_z=0$, the incompressibility condition. In the leading thin-sheet approximation, vanishing tangential traction makes $u$ independent of $z$. The normal stress boundary condition is ${\sigma }_{zz}=-p_{\mathrm{ext}}$, so the Newtonian fluid stress tensor givesThereforeFor the small annular sector, the inner and outer radial faces contribute $2\delta \theta {\partial }_r(rh{\sigma }_{rr})\delta r$ in the radial direction. The two azimuthal faces contribute $-2\delta \theta h{\sigma }_{\theta \theta }\delta r$: their hoop tractions have inward radial components. The combined radial force of the external pressure on the sloping upper and lower surfaces is $2\delta \theta r{p}_{\mathrm{ext}}{h}_r\delta r$.

Neglecting inertia, force balance is thusSubstituting the two stresses cancels the terms involving $p_{\mathrm{ext}}{h}_r$ and gives the axisymmetric viscous-sheet stretching equations:Finally, conservation of mass in the sector gives

Because $\varphi$ is a conserved composition density, an infinitesimal material displacement $\mathbf{r}\mapsto \mathbf{r}+\mathbf{u}$ changes it at fixed position byThe second term is essential when the deformation is compressible. By the definition of the chemical potential, and taking $\mathbf{u}$ to vanish on the boundary,The same free-energy change written in terms of the stress tensor isSince $\mathbf{u}$ is arbitrary,This is the Korteweg force density of a diffuse-interface mixture.

The affine linear PTT fluid considered here uses a symmetric dimensionless tensor $A$ with $\stackrel{\nabla }{A}+(1+\alpha \mathrm{tr}A)A/\tau =2E$ and polymer stress $G_{0}A$. Here $\stackrel{\nabla }{A}$ is the upper-convected derivative, $G_0,\tau >0$ and the nonlinear relaxation parameter $\alpha$ is usually positive. Linearizing about rest gives the Maxwell fluid relaxation modulus $G(s)=G_0{e}^{-s/\tau }$. The PTT family also admits other relaxation functions and nonaffine derivatives; their nonlinear predictions need not equal those of this specified version.

These stress measures refer force or area to the reference configuration. Different index-order conventions distinguish a first Piola-Kirchhoff stress tensor from its transposed nominal stress tensor. The second Piola-Kirchhoff stress tensor has two material indices.

For fixed plates $y=\pm h$, no slip, viscosity ${\mu }_0>0$, yield stress ${\tau }_0\ge 0$ and $G=-p_{,x}>0$, force balance gives shear stress $-Gy$. The central plug flow of a yield-stress fluid has half-width $y_0={\tau }_0/G$. If $y_0<h$, the velocity isIt is even in $y$; its derivative is $(-Gy+{\tau }_0\mathrm{sign}y)/{\mu }_0$ in the yielded layers and zero in the plug. If $Gh\le {\tau }_0$, there is no flow. Odd velocity is incompatible with this pressure-driven no-slip solution at nonzero flow.

In plane strain, the deformation map has $x_1=x_1(X_1,X_2)$, $x_2=x_2(X_1,X_2)$ and $x_3=X_3$. The out-of-plane stretch is one and out-of-plane shear vanishes. This kinematic condition need not make the out-of-plane stress vanish.

Pressure measures the compressive normal force per unit surface area. For a uniformly loaded surface of surface area $A$, the magnitude of the perpendicular force is $F_{\perp }=PA$; locally, $P=d{F}_{\perp }/dA$. Its SI unit is the pascal, $1\mathrm{Pa}=1\mathrm{N}{\mathrm{m}}^{-2}$. A force parallel to the surface contributes shear stress, rather than pressure. In a gas, the steady macroscopic force comes from many microscopic transfers of momentum; ideal gas pressure from wall collisions makes this connection explicit. ^[ref]

For a computed example, a uniform pressure of $100\mathrm{kPa}$ on a piston of surface area $10\mathrm{c}{\mathrm{m}}^2=10^{-3}{\mathrm{m}}^2$ produces a perpendicular force of $100\mathrm{N}$. If the opposite face is exposed to $80\mathrm{kPa}$, the net force due to the two pressures is $(100-80)\times 10^3\times 10^{-3}=20\mathrm{N}$, directed toward the lower-pressure side. The pressure on either face and their difference answer different questions; gauge pressure records a difference from a specified reference, often atmospheric pressure.

A scalar pressure describes a gas in local thermodynamic equilibrium with isotropic random particle velocities: every wall orientation has the same normal force per unit surface area. Without this isotropy, normal stresses can depend on direction and a pressure tensor is needed. Thus random motion alone is insufficient; a preferred direction matters.

In thermodynamics, pressure and volume are thermodynamic conjugate variables. For a simple compressible system, reversible expansion delivers work $\delta W=PdV$, and at fixed entropy $S$ and particle number $N$,where $U$ is internal energy. A large pressure therefore means both a large wall force per unit surface area and a large amount of expansion work per small increase in volume.