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In a steady axisymmetric outflow, mass conservation makes $2\pi r\Sigma (r)v_r(r)$ independent of $r$. The steady radial dust outflow therefore hasIt has an integrable $(r-a_p{)}^{-1/2}$ pile-up just outside the source ring, where the radial speed starts from zero, and approaches $Sigma\propto r^{-1}$ at large radius.

For a homologous star, hydrostatic equilibrium, mass conservation, and the ideal gas equation give the central scalingswhere $\mu$ is the mean molecular weight. Integrating the nuclear energy-generation law over a fixed homologous profile givesRadiative stellar structure gives independentlyEquating the two luminosities yieldsAt fixed zero-age composition, the stars are therefore homologous with

The effective temperature satisfies $L=4\pi R^2\sigma T_e^4$, so $T_e\propto M^{3/8}$. The zero-age main sequence consequently hasIt is a steep line rising toward high luminosity and high temperature on a Hertzsprung-Russell diagram.

For fully ionized hydrogen and helium with $Y=1-X$,while $\kappa \propto 1+X$. At fixed mass,and the corresponding radius relation is

At the pure-hydrogen zero-age point $X=1$,whereasHenceandAs hydrogen is consumed, $X$ falls, so both $L$ and $T_e$ rise. In the usual diagram with temperature increasing leftward, the evolutionary track initially moves upward and leftward from the zero-age main sequence.

Let $z$ increase downward. The slender viscous thread is locally in uniaxial extension. The transverse stress equals the ambient pressure, so the Trouton ratio gives the excess axial stress ${\sigma }_{zz}+p_e=3\mu w_z$. Mass conservation and axial force balance therefore give

In steady flow $Aw=Q$. Dividing the momentum equation by $A=Q/w$ givesWith $w=\hat{w}W$ and $z=\hat{z}Z$, chooseThe dimensionless equation becomesTreating $(W^{\prime }{)}^2$ as a function of $W$ and using an integrating factor yieldsBecause $A=Q/(\hat{w}W)$, the area sketches are the reciprocals of the functions $W$ found below. The dimensional vertical deviatoric stress is

The broad faces have curvature zero to leading order, so their stress boundary condition gives ${\sigma }_{xx}=-p_e$. The semicircular edges have curvature $1/h$, giving ${\sigma }_{yy}=-p_e-\gamma /h$. Incompressible flow gives $u_x+v_y+w_z=0$; uniform transverse normal stresses imply uniform transverse extension rates, and eliminating them from the Newtonian fluid stress tensor yields

At an edge, the kinematic boundary condition balances axial advection of $b$, lateral strain, and capillary retraction. This givesThe equation $hbw=Q$ expresses conservation of volume flux, whilestates that the total axial tension is constant. DefineThen $[\Gamma ]=L^{-2}$ and $[T]=L^{-1}$, and substitution of $h=Q/(bw)$ givesSubtracting these logarithmic-derivative equations givesThe remaining width equation isThus, for $T\ne 0$,while the continuous $T=0$ limit is $1/b=1/b_0+\Gamma z/6$.

If $\Gamma =0$, then $b/b_0=e^{-Tz/6}$, $w/w(0)=e^{Tz/3}$, and mass conservation gives $h/h(0)=e^{-Tz/6}$. Therefore a prescribed thinning ratio $R=h(L)/h(0)$ requires

Let $K=k\Delta \rho g/\mu$. Hydrostatic pressure continuity at the lower boundary of the light current givesinside the current, so $p_x=\Delta \rho g{h}_x$. Darcy law gives the depth-integrated horizontal flux per unit transverse widthThus the porous gravity current satisfiesaway from the fracture. The boundary and front conditions areAt $x=L$, the pressure excess at the base of the fracture is $\Delta \rho g{h}_L$. Taking upward leakage as positive,This jump is the local mass conservation law for a leaky porous gravity current.

At late times, $q=Q$ to leading order on $0<x<L$, while $Q_l\simeq Q$. Integrating $q=-K(h^2{)}_x/2$ givesLeakage balance and the pressure-driven drop determine

In the far field, $h(L,t)\simeq h_L$ andBalancing the two sides with $h=O(h_L)$ givesMore precisely, setThe self-similar solution is determined byand $x_N=L+{\eta }_N(K{h}_{L}t/\varphi {)}^{1/2}$.

A shelfy stream is grounded ice with nearly depth-independent horizontal velocity. If the lubricating till has thickness $d$ and viscosity ${\mu }_t$, its simple-shear traction is approximately ${\mu }_{t}u/d$. Comparison with ${\tau }_b=\mu u/\lambda$ givesso $\lambda$ has dimensions of length.

Hydrostatic pressure contributes the depth-integrated longitudinal force $-\rho g{h}^2/2$, while the Newtonian extensional stress contributes $4\mu h{u}_x$ by the shallow-shelf approximation. Balancing the change of their sum against basal drag on a slice givesFor a freely floating ice shelf, hydrostatic flotation reduces the gravitational driving by $1-\rho /{\rho }_w$ and removes basal drag:Depth-integrated mass conservation isin the absence of accumulation or ablation.

In a thin stream with horizontal scale $L_x$, the ratio of extensional resistance to basal drag is $O(\lambda h/L_x^2)$, which is small when the thin-film condition $h/L_x\ll 1$ is combined with $\lambda \ll h_G$. For steady flux $hu=q$, neglecting extension givesAt the grounding line, flotation over a bed depth $b$ givesIntegration yieldsThe grounding-line slope is $|h_x(0)|=\mu q/(\rho g\lambda h_G^2)$. Thin-film theory therefore also requiresThis and $\lambda \ll h_G$ are compatible precisely when $\mu q/(\rho g{h}_G^3)\ll 1$.

The total horizontal force resultant in the stream isThe floating shelf equations and its ocean-front traction imply that the force it exerts at the grounding line is the ocean's hydrostatic force, soUsing flotation and the approximate stream solution,Consequentlyand the grounding-line flux-thickness relation is

At the final state of the slump test for yield stress, an axisymmetric deposit of height $h(r)$ is just able to support itself. Hydrostatic pressure gives radial pressure gradient $\rho g{h}^{\prime }(r)$, while lubrication theory makes the magnitude of the basal shear stressThe marginally yielded final profile therefore satisfiesIntegration givesUsing mass conservation, the known volume isSolving for the yield stress produces the estimateThus one measures the final radius $R$ and inserts it with $V$ and $\rho$. The estimate assumes a thin deposit, negligible surface tension, complete initial yielding, and a spatially uniform yield stress.

Dry sand is a granular material governed primarily by frictional stability. Its final free surface reaches the angle of repose ${\theta }_r$, so the deposit is approximately a cone,This constant-slope profile differs from the square-root edge of the yield-stress-fluid model.

On wavelengths much larger than halos, all matter is partitioned among halos. The mass-weighted halo overdensity must therefore equal the matter overdensity. Since ${\delta }_h(M)=b(M){\delta }_m$, mass conservation requires

For the Press-Schechter formalism, the mass-fraction measure becomesIt is normalized and is a half-normal distribution, soUsing the linear Eulerian halo biastherefore giveswhich verifies the consistency relation.

Since $Df/Dt=0$, the density ansatz hasThe velocity divergence isThe mass conservation equation therefore givesSimilarly,and hence

The $x$ component of the material acceleration isBecause $d\hat{p}/df=\hat{\rho }$,Combining this with $-{\partial }_x\Phi =-{\Omega }^{2}x$ givesThe $y$ and $z$ components give the analogous equations for $b$ and $c$. Finally, $\hat{p}(0)=0$ makes $p=0$ on the material free surface, so both dynamic and kinematic boundary condition requirements are satisfied.

The depth-integrated ice flux isLocal mass conservation with accumulation rate $a$ saysSubstituting the flux givesas required. Equivalently, after using $u_b=-\beta h_x$,

For $\mathbf{u}=\nabla \times (\psi \hat{\mathbf{y}})$,Hence $u_x+w_z=-{\psi }_{zx}+{\psi }_{xz}=0$, so mass conservation holds identically. Removing the hydrostatic part by writingand taking the curl of the Stokes flow equation gives the biharmonic equation

For a Fourier mode proportional to $e^{ikx}$ with $k>0$, decay as $z\to -\infty$ selectsThe velocity and pressure fields are thereforeDirect substitution verifies the momentum equation.

The horizontal velocities $u$ and $v$ point east and north, $\eta$ is the displacement of the free surface from its mean level, $H_0$ is the undisturbed depth, $g$ is gravitational acceleration, and $f$ is the constant Coriolis parameter on an f-plane. The three linearized shallow water equations are horizontal momentum balance and mass conservation:They follow from the rotating Navier-Stokes equation by assuming an inviscid homogeneous layer, hydrostatic pressure, horizontal scales much larger than $H_0$, depth-independent horizontal velocity, a flat impermeable bottom, constant $f$, and small surface displacement and velocity so that nonlinear products are neglected.

In a steady state, geostrophic balance givesThe relative vorticity is $\zeta =v_x-u_y$. Expanding the shallow-water potential vorticity $(f+\zeta )/(H_0+\eta )$ to first order givesThus one convenient normalization of its disturbance isTaking the curl of momentum and using continuity shows $q_t=0$.

Because $v={\psi }_x=(g/f_0){\eta }_x$, the mean boundary-current velocity is exactly related to the height change byFor the e-folding convention used in part d,Its robust narrow-layer scaling isThus weaker drag makes the current proportionally narrower and faster. Their product is independent of $\gamma$ to leading order:Mass conservation requires the narrow return transport to cancel the broad Sverdrup balance transport. The meridional gradient of planetary potential vorticity makes a frictional closure possible on the western side and produces western intensification; bottom drag supplies the vorticity sink that permits fluid parcels to cross potential-vorticity contours there.

The variables $u,v,w$ are eastward, northward, and upward velocity, $\varphi =p^{\prime }/{\rho }_0$ is the pressure perturbation divided by reference density, $\sigma$ is buoyancy, $N$ is the constant buoyancy frequency, and $\beta y$ is the equatorial approximation to the Coriolis parameter. The equations express, respectively:

Together they are the hydrostatic Boussinesq approximation for long equatorial waves.

Let $\bar{\rho }(t)$ be the homogeneous density. A Lagrangian coordinate volume $d^{3}q$ contains the same mass as its image $d^{3}x$ under the Zeldovich approximation, so mass conservation givesSince $1+\delta =\rho /\bar{\rho }$, the density contrast is

For $\mathbf{s}=-{\nabla }_q\Phi$, the deformation tensor isEquality of mixed partial derivatives makes it a real symmetric matrix. The real spectral theorem therefore supplies an orthonormal eigenbasis and real eigenvalues, which we denote by $-{\lambda }_1,-{\lambda }_2,-{\lambda }_3$. In that basis,and henceBefore the first crossing all factors are positive, allowing the absolute values to be omitted.

When $1-D{\lambda }_i=0$, the map loses rank in the corresponding principal direction. Its Jacobian determinant vanishes, trajectories meet, and shell crossing creates a cosmological caustic. The single-stream pressureless density formally diverges and the approximation no longer describes the subsequent multistream dynamics. If only one ${\lambda }_i$ is positive, one axis first collapses while the other two remain extended, producing a sheet or pancake. Collapse along a second and then a third principal axis produces filaments and nodes. Spatial variation of the eigenvalues joins these objects into the cosmic web around underdense voids.

The approximation succeeds because it reproduces linear growing-mode evolution exactly, preserves the initial tidal displacement and anisotropic collapse, and follows matter along nearly inertial comoving trajectories instead of expanding only the density at a fixed point. Large scales remain weakly nonlinear and are insensitive to the detailed dynamics after crossing. Its limitations begin at shell crossing: it permits streams to pass through one another, cannot produce virialized halos, and omits velocity dispersion, vorticity, gas pressure, shocks, feedback, and strongly nonlinear self-gravity.

For the one-dimensional displacement,so mass conservation givesThe earliest crossing occurs where the cosine is maximal:for integers $n$. The denominator vanishes at these isolated points.

Write $u=q-q_n$ at $t=t_c$. The Taylor series givesThus $|u|\sim (6|x-x_n|/k^2{)}^{1/3}$ andThe first caustic is therefore a cubic cusp with

For a material fluid volume, mass conservation yields ${\rho }_t+\nabla \cdot (\rho \mathbf{u})=0$. Constant mass density reduces this to $\nabla \cdot \mathbf{u}=0$. For inviscid flow, the Cauchy stress tensor is $-pI$; its net surface force is the volume integral of $-\nabla p$ by the divergence theorem. Newton's second law for each material volume therefore gives $\rho D\mathbf{u}/Dt=-\nabla p$. Since $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$, these are exactly the two stated equations.

In two dimensions they readDifferentiate the second momentum equation in $x$ and subtract the $y$ derivative of the first. The pressure terms cancel. For $\omega =v_x-u_y$, collecting terms givesIncompressibility removes the last term, givingThus each fluid parcel retains its scalar vorticity as it moves. In particular an initially irrotational flow remains irrotational in this smooth two-dimensional inviscid evolution. There is no vortex stretching in this two-dimensional setting.

For a fixed set of bodies in an inertial frame, total linear momentum is constant if the resultant external force vanishes. Summing Newton's second law cancels internal force pairs by Newton's third law. For an open subsystem, momentum carried across its boundary must also be included.

For an equal-intake-fuel mass-loading rocket, enclose the original craft and the stationary material collected during $dt$. Its initial momentum is $mu$. At the end, the craft has mass $m-\mu dt$ and the exhaust has mass $2\mu dt$. To first order,The first term is rocket thrust; the second is the momentum cost of collecting stationary matter. It vanishes in the conventional rocket equation, which therefore has no analogous fixed exhaust-speed ceiling in Newtonian mechanics. Here the acceleration vanishes at $u_{\max }=2u_0$; from rest it stays positive below that value while fuel remains. Relativistic limits are outside this model.

For constant $\mu$, mass conservation gives $m=m_i-\mu t$. Separating variables,ThereforeThe formal limiting speed would require $t=m_i/\mu$ and zero remaining craft mass. A positive rocket dry mass and finite carried fuel terminate the burn earlier. Maintaining nonzero collection rate even at rest is an idealization explicitly prescribed by the problem.

A convenient parameter is the water vapor mole fraction itself. The mass conservation calculation yieldsBoth functions increase for $0\le x\le 1$. At the endpoint,Therefore the physically admissible hydrogen–water vapor mixture requiresBeyond this bound, the formal positive quadratic equation root requires $x>1$, equivalently a negative hydrogen partial pressure. It is the physical constraints, rather than loss of real roots, that end the solution branch. For a concrete example, put $P_d=\Delta p=1$ in consistent pressure units, $ϵ=9$, and $E=3$. The positive root of $P^2+23P-51=0$ is about $2.037$, giving $x\simeq 1.473$, which is impossible.

The limiting temperature is obtained from $E(T_{\ast })=P_d+\Delta p$:This finite limit exists above $T_t$ if $E_t<P_d+\Delta p<E_t\exp [L_v/(R_v{T}_t)]$. The upper qualification matters mathematically: the stipulated constant-latent heat formula has a finite limiting saturation pressure as $T\to \infty$, so for an arbitrarily large unspecified $P_d+\Delta p$ the printed assertion of a finite breakdown temperature is not unconditional. For a concrete formal counterexample, choose $P_d+\Delta p=10^{15}\mathrm{Pa}$: with the usual triple-point constants the model has $E(\infty )\simeq 2.5\times 10^{11}\mathrm{Pa}$, so it never reaches the proposed inventory limit. This is a counterexample inside the stipulated equations, not a physically credible high-pressure ideal gas ocean. Actual liquid water also ceases to exist above its critical point, before such an extrapolation would make sense.

Physically, the imposed finite-pressure-thickness layer can hold no more than $\Delta p/g$ of water vapor per unit area. As $x\to 1$, it expels all its hydrogen into the dry upper layer. A still larger surface saturation pressure cannot coexist with this geometry. The layer must deepen, the temperature profile must change, or the assumed phase equilibrium must fail. If $P_d<\Delta p$, the additional restriction $P>\Delta p$ can also exclude the low-temperature end of the branch.

The first equation is the local balance of linear momentum. Use $\mathsf{M},\mathsf{L},\mathsf{T}$ for the dimensions of mass, length, and time, to distinguish dimensional time from the Cauchy stress tensor $\mathbf{T}$.

Here $t$ has dimension $\mathsf{T}$. The hypothesis ${\partial }_t\mathbf{b}=0$ means that the prescribed spatial body force field is the same for every member of the static family and the resulting motion.

Let $t(n)$ denote traction: the force per unit area exerted across an oriented internal surface with unit normal $n$. Define ${\sigma }_{ij}$ as component $i$ of the traction on a coordinate plane whose normal is $e_j$. Balance of linear momentum on a shrinking tetrahedron makes the surface-force terms, of order area, dominate its inertia and body forces, of order volume. Resolving its oblique face into the three coordinate projected areas gives Cauchy's traction formulaOpposite faces have opposite tractions, as follows independently from a shrinking pillbox. This fixes both the index and normal conventions.

For a material volume $V(t)$, the Reynolds transport theorem says $d{\int }_{V(t)}fdV/dt={\int }_{V(t)}[f_t+\nabla \cdot (fu)]dV$. Apply it componentwise to $f=\rho u_i$ and use mass conservation. With no body force in this calculation, balance of linear momentum and the divergence theorem giveArbitrariness of the material volume yields $\boxed{\rho D{u}_i/Dt={\partial }_j{\sigma }_{ij}}$, with $D/Dt={\partial }_t+u_j{\partial }_j$.

For constant dynamic viscosity and incompressible velocity, ${\partial }_j{u}_j=0$, soThus $\boxed{\rho (u_t+u\cdot \nabla u)=-\nabla p+\mu {\nabla }^{2}u}$.

In fully developed steady pipe flow $u=w(x,y)e_z$, the convective acceleration vanishes. The transverse equations give $p_x=p_y=0$ and the axial equation gives $p_z=\mu (w_{xx}+w_{yy})$. One side depends only on $z$ and the other only on $(x,y)$, so both equal a constant. HenceIf $n$ is outward from the fluid cross-section, the wall-on-fluid axial shear per length is $\mu \oint {\partial }_{n}wds=\mu {\int }_A\Delta wdA=A{p}_z$. The fluid-on-pipe force is its negative:For forward pressure-driven flow $p_z<0$, this force points along the flow. Its value depends on cross-sectional area and pressure drop, even though the detailed velocity profile depends on the shape.