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Choose ${\rho }_0$ as a representative room density, for example the fresh-air density, and neglect relative density variations everywhere except in buoyancy. Withthe triangular profiles giveSolving these algebraic relations,

The Batchelor entrainment hypothesis, vertical momentum conservation, and mass conservation giveAn ascending parcel entrains ambient fluid from progressively lower ambient density. With the buoyancy frequencythe change of ambient reference density subtracts $Q{N}^2$ from its density-weighted buoyancy flux, so

At radius $r$, radiation of luminosity $L$ gives matter of flux-mean opacity $\kappa$ the outward accelerationEquating this with the gravitational acceleration $G{M}_{\mathrm{BH}}/r^2$ gives the Eddington luminosity

Let $L=f{L}_{\mathrm{Edd}}$ and let $\mathcal{M}(r)$ be the radiative force multiplier relative to the Compton or electron-scattering acceleration. A steady spherical wind obeys mass conservation ${\dot{M}}_w=4\pi r^2\rho v$ and the radial Euler momentum equationWhen gas pressure is negligible, radiation accelerates the wind locally if $f\mathcal{M}>1$. Electron scattering alone has $\mathcal{M}\simeq 1$, so it launches a wind only near or above the Eddington ratio $f=1$. Bound-free absorption, spectral lines, or dust can give $\mathcal{M}\gg 1$ and allow a sub-Eddington source to launch a wind, provided the gas is not so highly ionized that those extra opacities disappear. A wind that has already acquired sufficient kinetic energy may coast through a region where $f\mathcal{M}<1$, but such a region cannot launch a cold wind from rest.

Write the innermost stable circular orbit asEstimating the emitting area as $4\pi R_{\mathrm{ISCO}}^2$, the Stefan–Boltzmann law givesand henceThus $T_{\mathrm{BB}}\propto M_{\mathrm{BH}}^{-1/4}$ at fixed $f$ and opacity: a larger black hole radiates from an area growing as $M_{\mathrm{BH}}^2$, faster than its Eddington luminosity grows. Prograde black-hole spin moves the ISCO inward and raises the characteristic temperature, whereas retrograde spin moves it outward and lowers the temperature. Relativistic transfer and the radial temperature profile change the numerical coefficient, not these leading scalings.

In standard perturbation theory in cosmology, ${\delta }^{(n)}$ is an $n$-leg vertex carrying the symmetrized standard perturbation theory density kernel $F_n$. For Gaussian random fields, a connected four-point diagram with $L$ loops contains $3+L$ linear power spectra, hence $6+2L$ linear fields.

At tree level the required perturbative-order partitions areThe first is a star with one $F_3$ vertex and three linear external legs. The second has two $F_2$ vertices joined by one internal power-spectrum line, with one linear external leg attached to each vertex. At one loop all five partitions of eight into four positive parts are required:These are conventionally denoted $T_{5111}$, $T_{4211}$, $T_{3311}$, $T_{3221}$, and $T_{2222}$; all inequivalent placements of external legs and all Wick contractions are understood. This list is the complete set of one-loop matter trispectrum topologies.

Let $P_a=P_L(k_a)$ and ${\mathbf{k}}_{ab}={\mathbf{k}}_a+{\mathbf{k}}_b$. With ${\sum }_a{\mathbf{k}}_a=0$, the star contribution iswhere $\{a,b,c\}=\{1,2,3,4\}\setminus \{d\}$. For the exchange contribution, sum over the six unordered choices $\{a,b\}$ of linear external legs and let $\{c,d\}$ be the complementary pair:Thus the connected tree trispectrum is $T_{\mathrm{tree}}=T_{3111}+T_{2211}$.

The pressureless single-stream equations cease to be a complete one-loop prediction because loop momenta probe short nonlinear scales and produce ultraviolet-sensitive terms. A consistent result must include the effective field theory of large-scale structure: effective-stress counterterms, including their insertions into the tree topologies, and stochastic terms at the order allowed by mass conservation and momentum conservation. Their coefficients absorb short-scale dependence and must be fitted or matched. If the observable is a biased tracer rather than matter itself, the corresponding renormalized bias expansion is also required.

Steady spherical mass conservation gives $4\pi r^2\rho u=\dot{M}$. For the polytropic equation of state, $c_s^2=dp/d\rho$, soSubstitution into the radial Euler equations for an inviscid fluid gives the Parker wind equationAt a sonic point, the coefficient of $du/dr$ vanishes. A smooth transonic branch can pass it only if the right-hand side vanishes simultaneously. Hence

Write the perturbations as $(\sigma ,u,v)e^{i{k}_{x}x+\lambda t}$, set $k=|k_x|$, and evaluate all unmarked background quantities at ${\Sigma }_0$. DefineThe linearized mass conservation equation isThe radial and azimuthal momentum equations arewhere the razor-thin disk Poisson kernel gives $\varphi =-2\pi G\sigma /k$. The term proportional to $\beta$ comes from perturbing the density-dependent background shear stress $T_{xy}=-\nu \Sigma S$.

Eliminating $\sigma ,u,v$ and using $S=3\Omega /2$, so that the radial epicyclic frequency obeys ${\kappa }_r^2=2\Omega (2\Omega -S)={\Omega }^2$, gives the dispersion relation

Let $\zeta =z/H(t)$ label a fluid element and use the homologous vertical motion of an astrophysical diskBecause $D\zeta /Dt=0$ and $\nabla \cdot \mathbf{u}=\dot{H}/H$, this ansatz satisfies mass conservation. The pressure equation for an adiabatic process giveshenceThe vertical acceleration is $D{u}_z/Dt=\ddot{H}\zeta$. Using the dimensionless hydrostatic profiles from part c, the pressure force is $(P/\Sigma H)\zeta$, and the vertical momentum equation reduces towhere $C$ is constant.

The long-wave approximation makes $u$ independent of $z$ at leading order. Mass conservation and symmetry about $z=0$ then giveFor an incompressible Newtonian fluid,The leading normal-stress balance on either surface isSince ${\sigma }_{xx}-{\sigma }_{zz}=4\mu u_x$, it follows that

For a slice of length $\delta x$, the axial forces are the integrated normal stresses $2h{\sigma }_{xx}$ on its vertical ends, ambient pressure on the varying end height, and the horizontal components of surface tension on its two sloping faces. Expanding their difference to first order in $\delta x$ cancels the uniform $p_a$ terms and the lower-order capillary terms, leavingThe kinematic condition on $z=h(x,t)$ is $h_t+u{h}_x=w$. Substituting $w=-h{u}_x$ gives the second required relation,

Take $x$ downslope, $y$ across the slope, and $z$ normal to the plane. The leading normal momentum balance givesThe tangential lubrication theory equations, with no slip at $z=0$ and zero tangential stress at $z=h$, then give the depth-integrated fluxFor $\alpha \ll 1$, defineDropping tildes, steady mass conservation $\nabla \cdot \mathbf{q}=0$ becomeswith $h\to 1$ as $x\to -\infty$.

Let $u=-v_r>0$ denote the inward speed. Steady spherical mass conservation and the radial momentum equation for an isothermal gas giveEliminating $d\log \rho /dr$ produces the Isothermal Bondi equationA smooth flow can cross Mach number one only where both sides vanish. Its Bondi sonic point is thereforeThe integrated Bernoulli equation which approaches rest and density ${\rho }_{\infty }$ at infinity isAt the sonic point this gives ${\rho }_s=e^{3/2}{\rho }_{\infty }$, and hence the unique regular transonic branch has

Steady mass conservation gives $\dot{M}=-2\pi R\Sigma u_R$. Substituting this into the angular-momentum equation and integrating from the innermost stable circular orbit with zero torque givesThereforeUsing $\nu =\alpha c_{s}H$, $c_s\simeq H\Omega$, $l=R{u}_{\varphi }$, and $Rd\Omega /dR$ of order $-\Omega$ givesup to the order-unity Keplerian factor $3/2$.

At the sonic transition, $|u_R|\simeq c_s\simeq (H/R)u_{\varphi }$. Hencefor a geometrically thin disk with $\alpha \ll 1$. The specific angular momentum therefore differs only fractionally from $l_{\mathrm{ISCO}}$ before the gas enters the plunging region of a black-hole accretion disk. Its much shorter inflow time then prevents appreciable viscous transport, justifying angular-momentum conservation across the ISCO and the zero-torque boundary condition.

For an adiabatic Bondi accretion flow, $\dot{M}\propto M^2{\rho }_{\infty }{c}_{s,\infty }^{-3}$. Inside the Bondi radius the speed and ion temperature are approximately virial, so $v\propto (M/r{)}^{1/2}$ and $T\propto M/r$, while mass conservation gives $n\propto \dot{M}{M}^{-1/2}{r}^{-3/2}$. The frequency-integrated thermal bremsstrahlung emissivity is proportional to $n^2{T}^{1/2}$. Its volume integral is dominated by the inner flow and consequently scales asSince the Eddington luminosity is proportional to $M$, this may be written $L/L_{\mathrm{Edd}}=C(\dot{M}{c}^2/L_{\mathrm{Edd}}{)}^2$. With the standard fully ionized-plasma constants and the Bondi profiles, $\sqrt{C}=9\times 10^{-3}$. Eliminating $\dot{M}$ from $\eta =L/(\dot{M}{c}^2)$ then givesThe cancellation of $M$, ${\rho }_{\infty }$, and $c_{s,\infty }$ expresses the scale-free character of the ideal flow. More physically, two-body emission scales as density squared, so an increasingly dilute flow radiates a progressively smaller fraction of its available accretion power.

Let $u(r)>0$ denote the inward radial velocity. Steady spherical symmetry and mass conservation giveThe radial Euler momentum equation for the globally isothermal equation of state $p=c_s^2\rho$ isThe logarithmic derivative of mass conservation is ${\rho }^{\prime }/\rho =-u^{\prime }/u-2/r$. Substitution gives the Isothermal Bondi equationAt a smooth sonic point, $u=c_s$ makes the coefficient of $u^{\prime }$ vanish, so the right-hand side must vanish too. ThereforeThe solution that crosses this critical point of the isothermal Bondi equation continuously is the transonic accretion solution.

With the dimensionless variables $x=r/r_s$ and $y=u/c_s$, mass conservation and part (c) giveSubstitution into the Bernoulli function, together with $GM/r=2c_s^2/x$, eliminates every dimensional parameter and gives the transcendental equationEquivalently,orThe transonic branch passes through $(x,y)=(1,1)$.

For $u_0>u_A$, define the Alfvénic Mach-cone slopeThe equation in part (b) is the hyperbolic partial differential equationwhose characteristic curves are $x\pm mz=\text{constant}$. Let $W(s)$ be one for $|s|<L/2$ and zero otherwise. Selecting the characteristics that trail downstream, toward negative $x$, and imposing part (c) givesThe stationary induction and mass conservation equations then givewith $\delta u_z=\delta u_y=\delta B_y=0$. In the strongly super-Alfvénic limit, $\delta u_x\simeq u_{A}b/B_0$ and $\delta B_z\simeq b{u}_A/u_0$ inside the disturbed region.

The perturbations have support only whereBecause the wing is infinite in $y$, this is the union of two inclined slabs bounded by the characteristic planes $x+m|z|=\pm L/2$. These two trailing slabs are the Alfvén wings generated by the conductor.

The Strong-shock Rankine-Hugoniot conditions give the compression ratiofor $\gamma =5/3$. ThusIn the shock frame, mass conservation saysUsing ${\rho }_2=4{\rho }_0$ gives

For a family in stellar homology, dimensionless radial profiles are fixed. Mass conservation and hydrostatic equilibrium then give the central scalingsThe ideal gas equation of state, with fixed mean molecular weight, consequently giveswhere fixed dimensional constants such as $G$ are suppressed in homology relations.

The proton–proton chain law $ϵ\propto \rho T^{7/2}$ gives the nuclear luminosityOn the other hand, radiative diffusion in a star givesUsing the Kramers' opacity law ${\kappa }_c\propto {\rho }_c{T}_c^{-7/2}$,Thermal equilibrium requires $L_{\mathrm{nuc}}=L_{\mathrm{rad}}$. Their common mass factor cancels, leaving $R^{-13/2}\propto R^{-1/2}$, and hence

This model captures the gas-pressure support, pp-chain burning, and strongly mass-dependent luminosity of the lower main sequence, including the Sun approximately. Its fully radiative assumption is an idealization: the Sun has a convective envelope, and sufficiently low-mass red dwarfs become largely or fully convective, so real radii are not exactly constant.

Define the surface density of a disk, outward mass flux, internal torque, and surface magnetic torque byThe sign convention makes $\mathcal{G}$ positive for outward angular momentum transport in an ordinary Keplerian accretion disk. Vertical integration of mass conservation gives

The specific angular momentum is $h=r{u}_{\varphi }=r^2\Omega$. Multiply the azimuthal equation by $r$, use the continuity equation to put its left-hand side in conservative form, and integrate over $z$. The assumed decay removes the vertical mass and viscous fluxes, whereas the magnetic surface stress remains:Subtracting $h$ times the integrated mass equation yieldsSince $F_M=-(dh/dr{)}^{-1}({\partial }_r\mathcal{G}-r\mathcal{T})$, substitution in mass conservation gives the required one-dimensional advection-diffusion equation

Axisymmetric mass conservation in the spherical gap givesSince $h_t=-V\cos \theta$ and regularity requires $q\sin \theta =0$ at $\theta =0$, integration gives

The leading pressure-driven lubrication flux isSubstitution of $h$ and $q$ givesand therefore

Take downward as the positive vertical direction. The constant pressure contributes no resultant, while the pressure force on the inner sphere is opposite its outward normal. ThusWith $t=\lambda \cos \theta$ and the supplied integral,The sign is upward for $V>0$, so this is a drag force. As $\lambda \to 0$, the bracket is $4{\lambda }^3/3+O({\lambda }^5)$ and $F_z\to -8\pi \mu a^{4}V/{\Delta }^3$.

The condition $a\gamma =a_0{\gamma }_0$ makes the leading extensional axial force uniform, but it does not make the capillary pressure uniform. Sincea nonuniform radius gives a nonuniform pressure. Its axial gradient must drive flow, so the prediction $w=0$ is inconsistent.

For $a=a_0+\eta$ varying over the axial scale $L$,An axial Hagen-Poiseuille flow in a cylinder has speed scaleCross-sectional mass conservation gives ${\eta }_t\sim a_{0}w/L$, and consequently

Let $z$ measure height above the sponge. Under the long-wave approximation, hydrostatic pressure gives $p_x=\rho g{H}_x$, and the horizontal balance for viscous fluid flow isThe no-slip boundary condition $u(0)=0$ and stress-free boundary condition $u_z(H)=0$ giveIntegrating this lubrication theory profile gives the horizontal volume flux per unit span

Local mass conservation includes the downward loss found in part i:Consequently the required nonlinear partial differential equation isIf the imposed inlet flux is $Q_0{t}^{\alpha }$ and the moving front is $x=x_N(t)$, sufficient boundary conditions areFor a current released onto a dry substrate one also takes the initial condition $H(x,0)=0$ away from the source. The front position is part of this moving-boundary problem.