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Use the Dupuit approximation: the saturated region is shallow enough that hydrostatic pressure is $p=\rho g(h-z)$ and flow is predominantly horizontal. Darcy's law then gives the horizontal Darcy velocity $u_x=-u_b(1+\beta z)h_x$, where $u_b=k_0\rho g/\mu$. Integrating through the saturated depth gives the volume flux per unit widthThe stored water volume per unit area is $\varphi h$, so mass conservation yields the unconfined aquifer with depth-dependent permeability equationDefine the positive river discharge by $Q(t)=-q(0,t)$. A useful check on all subsequent results is the integrated mass conservation law

Let a representative fixed volume have mixture mass density $\bar{\rho }={\rho }_s\phi +{\rho }_l(1-\phi )$. Since the crystal framework is rigid and stationary, the liquid mass flux is ${\rho }_l\mathbf{u}$, where $\mathbf{u}$ is Darcy velocity. Mass conservation givesThe constant phase densities therefore imply the phase-change volume source in a rigid mushFor ice less dense than brine, increasing solid fraction creates an expansion flow; when the phase densities agree, this source vanishes.

Use the common-density, equal-volume-heat-capacity model implicit in the stated speed law. Let the liquid fraction be $\varphi$ ahead of the melting front and $\varphi +\Delta \varphi$ behind it. Melting adds exactly the liquid needed to fill the newly created pores, so total mass conservation makes Darcy flux continuous. Consequently,Here $u$ denotes Darcy velocity and $v$ the pore-liquid velocity in the stationary rock frame. Treating the pore-space increase as storage without its simultaneous melting source would incorrectly change the Darcy flux.

Relative to the cold unmelted material, the bulk enthalpy increase behind the front is $\rho c_p\Delta T+\rho L\Delta \varphi$: all phases gain sensible heat and the melted ice consumes latent heat. The advective heat-flux difference is $\rho c_{p}U\Delta T$. Applying the Rankine-Hugoniot condition to this energy balance therefore givesso the advection-driven melting front in a porous matrix moves atThe advected latent contribution of the liquid is the same on both sides and cancels; the denominator measures sensible heating plus phase-change energy per unit bulk volume.

Let $\ell =(H_m-H_0)/\sin \alpha$ be the along-slope distance to the snowline, so $H(x)=H_m-x\sin \alpha$ and $a(x)=A(1-x/\ell )$. Define $D=g\sin \alpha /(3\nu )$.

Treat ice as an incompressible Newtonian fluid, neglect inertia, and use lubrication theory with thickness measured normal to the slope. The bed has a no-slip boundary condition, the free surface has zero tangential stress, and normal pressure is hydrostatic. The small thickness slope allows its pressure-gradient contribution to be neglected against gravity along the mountain. With normal coordinate $y$, the tangential equation and boundary conditions areIntegration givesThe horizontal radius of a ring is $x\cos \alpha$, so mass conservation gives the gravity-driven ice flow on a conical slope equationNegative accumulation is ice ablation and applies only where ice exists; the ice-free region has $h=0$.

In a steady state, regularity and zero total flux at the apex require $xD{h}^3\to 0$ as $x\to 0$. Integrating givesRequiring a continuous zero-thickness steady terminus yields the steady conical ice cap with a linear accumulation gradient:The terminus lies below the snowline, allowing the ablation region to balance snowfall. The maximum thickness occurs at $x=3\ell /4$.

The volume follows from integrating the ring areas. With $s=x/\ell$,Substituting $D$ and $\ell =\Delta H/\sin \alpha$ givesThe ideal outer profile has steep slopes very close to the apex and terminus. Those small regions require local corrections to the assumed slope balance, while the bulk profile and leading volume follow from the stated approximation.

With both ice accumulation and ice ablation removed, the equation is $h_t+x^{-1}(xD{h}^3{)}_x=0$ and the conserved volume is $V_0=2\pi \cos \alpha M$, where $M={\int }_0^{x_N}xhdx$.

If the thickness and extent scales are $H(t)$ and $X(t)$, mass conservation gives $H{X}^2\sim M$, while the flow equation gives $H/t\sim D{H}^3/X$. Therefore $X\propto t^{1/5}$ and $H\propto t^{-2/5}$. Set $h={\tau }^{-2/5}f(\xi )$, $\xi =x{\tau }^{-1/5}$, with a possible virtual time origin $\tau =t+t_0$. The similarity solution satisfiesIntegration and regular zero total flux at the apex give $D\xi f^3={\xi }^{2}f/5$, hence the positive profile iswith dry bed beyond the front. Volume normalization givesso the volume-conserving conical ice-current similarity solution hasThe terminus has finite thickness and is a shock wave in the gravity-only kinematic wave equation. The Rankine-Hugoniot condition gives ${\dot{x}}_N=q_N/h_N=D{h}_N^2=x_N/(5\tau )$, exactly agreeing with the similarity extent. The characteristic speed behind the front is $3D{h}_N^2$, larger than its speed, while the dry-bed characteristic speed is zero, so the front is compressive and gives an entropy solution.

This solution describes the long-time spreading, rather than exactly matching the earlier steady profile at the instant snowfall stops. The initial transient can be described by the characteristic transformation for conical ice drainage: with starting point $s$, put $K(s)=s^{1/3}{h}_s(s)$; thenwhere characteristics remain smooth. Subsequent crossings are resolved by the same conservation and entropy conditions. Restoring the neglected local pressure gradient would smooth the idealized front.

Initially, away from the apex and snowline, transport is small and local ice accumulation dominates:The early ice volume is $V(t)\simeq \pi A{\ell }^2\cos \alpha t/3$. Balancing radial transport $D(At{)}^3/\ell$ with $A$ identifies the global transition scalesThere is also an earlier local adjustment near the apex: the factor $1/x$ makes the same flux significant at distance $x_a\sim D{A}^2{t}^3$. The apex filling layer of a conical ice cap is described byAt the apex, $F\sim (\eta /2{)}^{1/3}$ gives the local steady profile $h\sim (Ax/(2D){)}^{1/3}$. Thus gravity-driven redistribution reaches a fixed position $x$ on a time of order $[x/(D{A}^2){]}^{1/3}$, and its region grows out to the snowline when $t\sim t_{\ast }$.

At times comparable to $t_{\ast }$, ice transport becomes important across the mountain's accumulation region, carries ice below the snowline, and establishes an ablation zone. The cap then approaches the steady conical ice cap with a linear accumulation gradient, with $x_N\to 3\ell /2$. For a moving ice-covered disk, mass conservation and its front condition giveThe total input vanishes at $x_N=3\ell /2$, explaining the final extent. The narrow apex and front regions are governed by local adjustments to the bulk approximation; they do not create a further distinct large-scale time regime in this scaling description.

Assume nonrelativistic pressureless matter, an initial irrotational vector field of growing-mode linear cosmological density perturbations, a homogeneous expanding background, and no shell crossing. All gradients below refer to initial comoving coordinates. Let ${\rho }_0={\bar{\rho }}_m(t)a^3(t)={\bar{\rho }}_m(t_i)$, with $a(t_i)=1$, and normalize the linear growth factor by $D(t_i)=1$. The paper calls $D$ a growth rate, but it is the dimensionless growth factor; $\dot{D}$ is its time derivative.

Writing ${\mathbf{x}}_i$ for the actual position at $t_i$, the Zeldovich approximation is, to first order,This explicitly gives $\mathbf{x}(t_i)={\mathbf{x}}_i$. The usual notation $\mathbf{x}=\mathbf{q}+D\mathbf{s}(\mathbf{q})$ instead uses the uniform reference Lagrangian coordinate $\mathbf{q}$, with ${\mathbf{x}}_i=\mathbf{q}+\mathbf{s}(\mathbf{q})$. The two labels can be interchanged inside a first-order displacement, but not in the unperturbed position without the initial displacement correction.

To determine $\mathbf{s}$, mass conservation between the initial and current positions givesDemanding $\delta (t)=D(t){\delta }_i$ gives $\nabla \cdot \mathbf{s}=-{\delta }_i$. The irrotational vector field assumption makes $\mathbf{s}$ a gradient; the initial cosmological Poisson equation is ${\nabla }^2{\Phi }_i=4\pi G{\rho }_0{\delta }_i$. With the same boundary conditions for these potential equations, and after removing an irrelevant uniform translation,The evolution of $D$ follows from the linearized peculiar-motion equation $\ddot{\mathbf{x}}+2H\dot{\mathbf{x}}=-\nabla \Phi /a^2$. Since cosmological Poisson equation gives $\Phi (t)=(D/a){\Phi }_i$ to this order, substitution yields the linear growth equationThus the approximation extrapolates the linear growing displacement along a fixed initial direction, even as its density mapping becomes nonlinear.

The early spin of a dark matter halo comes from an external gravitational torque on its nonspherical initial mass region. A uniform external acceleration moves its centre of mass without spinning it; the spatially varying tidal tensor exerts different forces on different parts. The tidal torque theory requires misalignment of the region's shape and the surrounding tidal field. A local irrotational vector field of velocity does not imply zero integrated angular momentum for a nonspherical region.

Put $\mathbf{y}={\mathbf{x}}_i-{\hat{\mathbf{x}}}_i$ and use the physical peculiar velocity $\mathbf{v}=a\dot{\mathbf{x}}$. The physical lever arm is $a(\mathbf{x}-\hat{\mathbf{x}})$, while a leading-order mass element is ${\rho }_0{d}^3{x}_i$. Substituting the velocity formula into the supplied definition of angular momentum about the centre of mass givesTerms from the perturbed mass measure are higher order. Subtracting the barycentre's velocity also changes nothing because ${\int }_{V_L}\mathbf{y}{d}^3{x}_i=0$ at this order.

Apply the divergence theorem componentwise:The antisymmetry of the Levi-Civita symbol kills the last term. Consequently, for the ordinary oriented comoving surface element $d\mathbf{S}$,The original PDF prints a different coefficient, $-\dot{D}/(4\pi G{\bar{\rho }}_m{a}^2)$. That coefficient does not follow from its own mass measure and velocity equation and has the wrong dimensions for total angular momentum. The corrected coefficient above also gives the time dependence consistent with the paper's later tensor expression; this is a source typo, not a TeX transcription issue.

For a spherical $V_L$ centered on its barycentre, the outward normal is parallel to $\mathbf{y}$, so $\mathbf{y}\times d\mathbf{S}=0$ pointwise. HenceFor an equipotential surface, ${\Phi }_i={\Phi }_{\ast }$ on ${\Sigma }_L$, andThereforeThe second conclusion holds for any boundary shape at this order, not just a sphere.

To extract the leading tidal torque, expand the initial peculiar gravitational potential about the barycentre using its Taylor series:The constant has no force; $\mathbf{g}$ has zero integrated torque because the first mass moment vanishes. With the mass second-moment tensorwe obtainThe second equality swaps the indices $k,l$ and uses the symmetry of both tensors. Define the initial acceleration tidal tensor byThen the formula in the question has exactly the stated sign:If one defines $T$ as the positive Hessian matrix of ${\Phi }_i$ instead, the displayed contraction has a plus sign. Defining the convention is essential. Also, the paper's $I$ is a mass second-moment tensor; the mechanical inertia tensor is $\mathrm{tr}(I)1-I$.

Only the anisotropic parts contribute: a multiple of the identity produces zero contraction with the Levi-Civita symbol. In axes where $I=\mathrm{diag}(I_1,I_2,I_3)$, for example,Thus a spherical region or tensors sharing principal axes have zero leading torque; unequal principal moments together with off-diagonal tidal components create spin. The Taylor series truncation assumes that higher spatial derivatives of the tidal field are sufficiently small across the region.

In an Einstein-de Sitter universe, $a\propto t^{2/3}$, $D\propto a$, and ${\bar{\rho }}_m\propto a^{-3}$. The initial $I$ and $T$ are time independent, soThis is early-time growth for fixed initial matter, before collapse invalidates the extrapolation. It does not predict indefinite linear spin growth for a virialized halo.

Two reasons for only approximate agreement with simulations are that nonlinear collapse and shell crossing change the trajectories and the tidal field, invalidating the first-order displacement and fixed leading tidal tensor; and that real halos undergo dark-matter halo mergers, anisotropic accretion, and exchange of material and angular momentum, so the halo identified at a later time need not be the same isolated collection of initial particles. These processes can change both the magnitude and the direction of the spin.

Take $u(r)>0$ to be the inward radial speed and select the regular critical solution for a gas reservoir of density ${\rho }_{\infty }$ and adiabatic sound speed $c_{\infty }$. Other mathematically subsonic steady solutions need a different inner boundary condition. For the polytropic equation of state, let $\gamma =1+1/n$, soMass conservation and the Bernoulli equation giveThe gas is at rest at infinity, so its kinetic and gravitational terms vanish there.

Differentiate these two conserved quantities, using $d\log \rho /dr=-2/r-u^{\prime }/u$. Eliminating the density derivative yields the polytropic Bondi accretion equationAt a regular sonic point, both sides vanish:The Bernoulli equation at that point reduces to $n{c}_{\infty }^2=(n-3/2)c_c^2$. For $n>3/2$, equivalently $1<\gamma <5/3$,Since $c_s^2\propto {\rho }^{1/n}$, the critical density is ${\rho }_c={\rho }_{\infty }[2n/(2n-3){]}^n$. Evaluate the conserved mass accretion rate at the sonic point:whereThe isothermal limit $\gamma \to 1$ gives $\lambda =e^{3/2}/4$, agreeing with Isothermal Bondi accretion.

For $\gamma =5/3$, $n=3/2$ and the Newtonian critical radius tends to zero while $c_c$ diverges. The limiting critical solution approaches Mach number one only as $r\to 0$; it has no finite-radius Newtonian sonic transition. This is a failure of the Newtonian inner approximation, not an obstruction to accretion onto a black hole. A physical Schwarzschild black hole has a finite event horizon, and the regular relativistic inflow becomes supersonic outside it.

The singular-looking factor in the rate has a finite limit. Writing $ϵ=n-3/2$, its logarithm is $ϵ\log [n/ϵ]\to 0$, so the critical Bondi accretion rate for gamma equals five thirds isThis is also the leading cold-reservoir limit of relativistic spherical accretion; an exact rate for relativistically hot gas requires the relativistic equations, not this Newtonian expression.

For completeness, the finite relativistic sonic point can be seen without assigning it an arbitrary inner radius. Let $b=c_s^2/c^2$ denote the relativistic squared sound speed and $g=\gamma -1$. For an isentropic relativistic gas with $p=K{\rho }_{\mathrm{rest}}^{\gamma }$, its dimensionless specific enthalpy is $h=g/(g-b)$. Conservation of rest-mass flux and relativistic Bernoulli energy gives the critical relationsFor $g=2/3$ and $c_{\infty }\ll c$, expansion gives $b_{\infty }\simeq (9/4)b_c^2$, henceHere the asymptotic relativistic and Newtonian sound speeds agree to leading order. This finite-radius transition explains why the gas can reach the horizon despite the Newtonian $r_c=0$ result.

Let $u(r,z)$ and $w(r,z)$ be the radial and vertical velocity components. For steady incompressible axisymmetric flow without swirl, mass conservation and the relevant cylindrical components of the Navier-Stokes equation are

The vertical scale is $w\sim V$. Incompressibility over radial scale $R$ and gap scale $h$ gives $u\sim VR/h=V/\epsilon$. In the radial equation, inertia is smaller than transverse viscous stress bywhile radial viscous derivatives are smaller than $u_{zz}$ by ${\epsilon }^2$. The pressure scale is $\mu V{R}^2/h^3$. In the vertical equation, both viscous and inertial terms are smaller than the corresponding cross-gap pressure-gradient scale by at least $O({\epsilon }^2)$. Thus the leading lubrication theory equations are

For a homologous stellar sequence, mass conservation, hydrostatic equilibrium, and the ideal gas equation giveIntegrating the given stellar energy-generation rate over a fixed dimensionless profile givesTherefore

Letbe the local isothermal sound speed squared. Steady spherical mass conservation givesThe momentum equation, with radiative acceleration represented by the radiation-pressure gradient, isRadiative diffusion in a star givesUsing $P_{\mathrm{rad}}/P_g=(1-\beta )/\beta$ andthis becomesSubstitution of $d{P}_g/dr=a_i^{2}d\rho /dr+\rho a_i^{2}d\log T/dr$ and the continuity equation yields the wind equation

Its topology is that of the Parker wind equation. The coefficient of $dv/dr$ vanishes at the sonic line $v=a_i$. A smooth transonic solution must pass through a critical point where the right-hand side also vanishes; generic subsonic solutions are breezes or turn back, while generic supersonic branches cannot be joined smoothly to a quasi-static stellar atmosphere.

At the critical point,The assumed inequality $a_i^2\ll G{m}_r/r$ makes the second factor positive and close to one. Since $0<\beta <1$,Thus acceleration through a regular sonic point requires radiation to cancel most, but not all, of the effective gravity and in particularIf the local luminosity reached or exceeded $L_{\mathrm{crit}}$ in this diffusion model, the numerator would have the wrong sign for the subsonic branch to cross the sonic line smoothly under the cold-wind assumption.

Take the rock at $z=0$, the till–ice interface at $z=h$, and the ice surface at $z=h+H$. The common leading hydrostatic pressure has horizontal gradientThe thin-film momentum equations areImpose no slip at the rock, continuity of velocity and shear stress at $z=h$, and zero shear at the ice surface. Expanding for $h/H\ll 1$ gives the till flux and ice fluxThe basal shear stress isApplying mass conservation separately to the ice and till, with erosion source $E{\tau }_0$, givesand

In a steady state $q_i=q_0$. Balancing the two terms in the till equation over length $L$ givesand henceindependently of $q_0$. Without substantial lubrication, the usual shallow-ice balance gives $H^4\sim \mu q_{0}L/(\rho g)$. ThereforeThe sheet is essentially unlubricated when $\boxed{\mathcal{E}\ll 1}$.

For $\mathcal{E}\gg 1$, the sliding term dominates the ice flux:Integrating the steady till equation from $x=0$, where $H=H_0$ and $h=0$, givesEliminating $H_x$ between these equations yieldsWith $F=H/H_0$ and $\xi =x/L$, a second integration giveswhere the physical branch decreases from $F(0)=1$ to $F(1)=0$. The length condition determinesUsing this relation in the till thickness givesThus $H$ decreases monotonically from $H_0$ to zero. The till starts at zero, rises to one interior maximum at $F=1/\sqrt{3}$, and returns to zero at the margin.

Well inside the Bondi accretion radius, pressure is negligible and the inflow approaches the free-fall speedSteady spherical mass conservation givesEstimating the outward optical depth across a radial scale $r$ with electron-scattering opacity $\kappa ={\sigma }_T/m_p$,The integral optical depth to infinity differs only by a factor of two in this free-fall approximation.

At the Schwarzschild radius, $\tau =1$ whenIf the conventional Eddington accretion rate is ${\dot{M}}_{\mathrm{Edd}}=L_{\mathrm{Edd}}/({\eta }_0{c}^2)$, then ${\dot{M}}_{\mathrm{crit}}=2{\eta }_0{\dot{M}}_{\mathrm{Edd}}$. This comparison depends on the reference efficiency ${\eta }_0$, while ${\dot{M}}_{\mathrm{crit}}$ itself follows directly from optical depth.

Writing $\dot{m}=\dot{M}/{\dot{M}}_{\mathrm{crit}}$ givesFor $\dot{m}>1$, the inner flow is optically thick and its photosphere lies at

The diffusion speed is $v_{\mathrm{diff}}\sim c/\tau$. Henceand equality defines the photon-trapping radiusInside $r_t$, inward advection outruns outward diffusion, so photon trapping in an accretion flow carries most released radiation into the black hole.

Only the binding energy released outside $r_t$ escapes efficiently. Its order of magnitude isThe radiative efficiency of black-hole accretion therefore decreases inversely with supply rate:Order-one coefficients depend on the optical-depth and inner-boundary conventions, but the luminosity saturation and $eta\propto {\dot{M}}^{-1}$ are robust.

The azimuthal momentum equation with $u_y=-3\Omega x/2$ isHence the slow radial drift isSubstitution into mass conservation, ${\partial }_t\Sigma +{\partial }_x(\Sigma u_x)=0$, gives the nonlinear diffusion equationIt describes slow axisymmetric viscous spreading while the velocity remains close to the imposed Kepler shear. It omits epicyclic and acoustic waves, rapid transients, self-gravity, nonaxisymmetric structure, and edge dynamics for which the assumed shear and secular force balance fail.

Inside the ring, $\nu \Sigma =A{\Sigma }^3=A{\sigma }^3(1-x^2/w^2{)}^{3/2}$. Part a therefore givesThe edge is material, so $\dot{w}=u_x(w)$ andDirect substitution of the profile into the diffusion equation giveshence $\sigma w={\sigma }_0{w}_0$. This is exactly mass conservation, sinceUsing $\sigma ={\sigma }_0{w}_0/w$ in the width equation and integrating,ThereforeIn particular, $w\propto t^{1/4}$ at late times.