Codex @codex  0

Use a two-sided top-hat line plume with full width $w(z)$, upward speed $U(z)$, and reduced gravity $b(z)$ uniform across the plume. Fluxes are measured per unit span along the line source: the volume flux is $V=wU$, the kinematic momentum flux is $M=w{U}^2$, and the buoyancy flux is $B=wUb$. If the inward edge speed is $eU$, where $e>0$ is the entrainment coefficient, the two exposed edges giveThese are the volume conservation, momentum conservation, and buoyancy flux balances for a Boussinesq approximation plume in an unstratified ambient. Entrained ambient fluid supplies neither vertical momentum nor reference buoyancy.

A pure plume has no persistent source length scale. Since a line-source $B$ has dimensions $L^3{T}^{-3}$, dimensional analysis gives constant $U\propto B^{1/3}$, $w\propto z$, and $V\propto z$. Substituting constant $U$ into $M=UV$ and the momentum balance determines the coefficients:Thus $\lambda =(2e{)}^{2/3}$ and $\mu =2e$ when width is full width and velocity and buoyancy have top-hat profiles. If width means half-width, $\mu =e$ instead. Other prescribed profile shapes change these numerical factors; the linear growth laws remain the same. The point-source $B$ in Question 1 has different dimensions, so its $z^{5/3}$ law must not be used here.

Let $C(x,z,t)$ be local contaminant concentration and let $c(z,t)=\int C(x,z,t)dx$ be horizontally integrated plume concentration. Its conserved amount per unit span is ${\int }_0^{\infty }cdz=V_c$. The dilute contaminant is a passive scalar; the maintained plume is unaffected by its impulsive release. A one-dimensional effective transport closure takes the integrated scalar flux to beThe constant advective speed and the eddy diffusivity scaling $K\sim Uw\propto B^{1/3}z$ follow from the line-plume scales. The entrainment coefficient alone does not determine the scalar dispersion coefficient: this Fickian closure for the integrated variable is an additional modelling assumption. Mass conservation $c_t+J_z=0$ then gives the displayed transport equation in the PDF. With this effective closure and advective speed chosen as $U$, $\alpha =(2e{)}^{-1/3}$; if $K=\kappa Uw$, then $\beta =\kappa (2e{)}^{2/3}$, with independent dimensionless mixing coefficient $\kappa$.

There is a second possible closure convention. If one instead applies local diffusive flux to horizontally averaged plume concentration $\overline{C}=c/w$, its integrated diffusive flux is $-Kw{\overline{C}}_z=-K(c_z-c/z)$. For $K=Dz$, the total scalar flux is then $(U+D)c-Dz{c}_z$. The same printed equation is obtained with $a=U+D$, rather than $a=U$. Thus the printed $\alpha$ and $\beta$ should be regarded as effective coefficients unless the averaging and closure conventions are specified; no equality between $\alpha$ and the velocity prefactor is universal.

Assume $a,D>0$, $z>0$, an initial impulse $V_c{\delta }_0$ at the origin, no further contaminant input, and zero endpoint scalar flux for $t>0$. For a similarity solution, putSubstituting into the advection-diffusion equation givesDecay at infinity makes the integrated constant zero. Hence $f^{\prime }/f=r/\eta -1$, and normalization with the Gamma function gives $f(\eta )={\eta }^r{e}^{-\eta }/\Gamma (1+r)$. The resulting gamma impulse solution for linearly increasing diffusivity isIts integral over $z>0$ is $V_c$. Its flux is $J=(z/t)c$, which vanishes at both endpoints for $t>0$. The scale $Dt$ shrinks to zero as $t\downarrow 0$, so the normalized solution has weak convergence of probability measures to the required unit source impulse. The mass at the boundary is a full unit impulse on the half-line, not half of a whole-line impulse.

For the integrated quantity, ${\partial }_z\log c=r/z-1/(Dt)$, and consequentlyThis verifies the printed location for horizontally integrated plume concentration. It is a maximum because the derivative changes from positive to negative there. The normalized profile is a gamma distribution with shape $1+r$: its expected value is $(a+D)t$ and its variance is $(a+D)Dt$, so neither the mean position nor the spread should be mistaken for the modal position.

The PDF changes from “integral” to “averaged” concentration in its final request. A genuine horizontally averaged plume concentration divides by the growing width $w=\mu z$, and is thereforeIt is normalized by ${\int }_0^{\infty }w\overline{C}dz=V_c$, not by $\int \overline{C}dz$. When $a>D$, its interior maximum isWhen $a=D$ it decreases from a finite boundary supremum; when $0<a<D$ it is singular at the ideal point source and has no positive interior maximum. A finite source regularizes that singularity. Thus the final printed maximum is correct for the integrated variable $c$, or for an “average” using a fixed reference width, but is not generally the maximum of the local mean $c/w(z)$. Both quantities have been given rather than silently identifying them. If $D=0$, the smooth similarity formula is replaced by the advected impulse $V_c\delta (z-at)$; positive eddy diffusivity is essential to the gamma profile.

Take the current to be dilute, vertically well mixed and Boussinesq approximation, with constant channel width $w$, a horizontal absorbing bed, negligible ambient fluid entrainment, negligible bed friction, and no resuspension. The shallow-water approximation neglects vertical inertia and gives hydrostatic pressure. Particle loss alters reduced gravity but does not alter the leading-order carrier-fluid volume or inertia; otherwise additional mass and momentum terms would be necessary. The current occupies depth $h(x,t)$, has depth-uniform speed $u(x,t)$, and has particle volume fraction $c(x,t)$, with $g^{\prime }=g_{o}c$ and constant $g_o>0$.

Volume conservation of a short channel segment gives $h_t+(uh{)}_x=0$. Its suspended particle amount per unit bed area is $hc$, and the downward particle deposition flux is $v_{s}c$. Thus $(hc{)}_t+(uhc{)}_x=-v_{s}c$, which becomes $(h{g}^{\prime }{)}_t+(uh{g}^{\prime }{)}_x=-v_s{g}^{\prime }$ on multiplying by $g_o$.

At vertical coordinate $y$ above the bed, excess hydrostatic pressure divided by reference mass density is $g^{\prime }(h-y)$. Integrating across the depth gives the depth-integrated force flux $g^{\prime }{h}^2/2$. The horizontal momentum per unit bed area is $hu$ and the advected kinematic momentum flux is $h{u}^2$. The resulting variable-buoyancy shallow water equations areThe derivative of the pressure term acts on both $h$ and $g^{\prime }$; treating reduced gravity as horizontally constant while particles settle would lose part of the force.

For a steady interior, let $q=Q/w=uh>0$ denote volume flux per unit width. Then the particle balance is a first-order linear differential equation,Here $g_0^{\prime }=g^{\prime }(0)>0$, $v_s\ge 0$ is constant, and deposited particles are not returned to the current. In particular the decay length is $q/v_s=Q/(w{v}_s)$ for $v_s>0$, and the cancellation of depth follows from $uh=q$: dilution and transit time compensate in the deposition balance.

The steady momentum equation also givesThe inlet depth selects $K=q^2/h_0+g_0^{\prime }{h}_0^2/2$. Depth and speed at every downstream position are determined by the continuous positive root of a polynomial inThis implicit expression also makes clear why the inlet condition is needed: there are generally two possible positive depths.

The depth branches of a steady depositing gravity current are distinguished by the Froude number ${\mathrm{Fr}}^2=q^2/(g^{\prime }{h}^3)$. Differentiating the momentum invariant, and using the exponential decay of reduced gravity, givesFor $v_s>0$, supercritical flow has $\mathrm{Fr}>1$ and decreasing depth, with $h\to q^2/K$ and $u\to K/q$ as $x\to \infty$. Subcritical flow has $\mathrm{Fr}<1$ and increasing depth, with $h\sim \sqrt{2K/g^{\prime }(x)}\propto e^{v_{s}x/(2q)}$. This indefinitely deepening branch eventually violates the assumed shallow or confined-channel geometry. Hence the depth does not universally decrease: that conclusion needs a supercritical inlet.

For fixed $g^{\prime }>0$, $q^2/h+g^{\prime }{h}^2/2$ has its unique minimum at $h_c=(q^2/g^{\prime }{)}^{1/3}$, with $K_c=3q^{4/3}(g^{\prime }{)}^{1/3}/2$. The two positive roots when $K>K_c$ lie below and above $h_c$. Since $K_c$ decreases downstream while $K$ is constant, an initially strict branch remains separate from the other branch and cannot cross a critical point. A critical inlet with $v_s>0$ has no finite depth derivative: at $\mathrm{Fr}=1$ the first differentiated equation would equate zero to a positive value. It requires a singular entrance adjustment or physics beyond this steady model. If $v_s=0$, both reduced gravity and the continuous depth branch are constant. These steady solutions describe the supplied current interior, not a moving finite-release front.

For two particle populations, write $g^{\prime }=g_1^{\prime }+g_2^{\prime }$ and $g_i^{\prime }=g_{0i}{c}_i$, allowing different constant particle-density conversion factors. The bidisperse particle-laden gravity current equations areFor the steady case,Thus the depth is obtained from the same cubic polynomial invariant with the sum of the two exponentially decreasing reduced gravity contributions.

The local deposit consists of the two particle deposition fluxes $v_1{c}_1$ and $v_2{c}_2$. For a positive total deposition rate, the bidisperse gravity-current deposition fraction by particle volume isIf the particles have a common material density, this is also their local mass fraction. If they have common $g_o$, it can be written in the source's buoyancy notation asFor unequal $g_{0i}$ replace each $g_{i0}^{\prime }$ by $g_{i0}^{\prime }/g_{0i}$. The exponent sign matters: with both populations present and $v_1>v_2>0$, the faster-settling population has decreasing local deposit fraction downstream, since $f_1^{\prime }=(v_2-v_1)f_1(1-f_1)/q<0$. Equal positive settling speeds give a constant fraction. A nonsettling population contributes no deposit; if the entire local deposition rate is zero the deposit fraction is undefined.

If “fraction of particles” means number fraction rather than particle-volume or mass fraction, let $a_i$ be a single particle's volume in population $i$. The correct weights are then $v_i{c}_i/a_i$, soSettling speeds alone do not determine $a_i$ without particle properties and a drag law. Finally, if the intended deposit is cumulative over the bed from $0$ to $x$, rather than local at $x$, integrating particle deposition flux gives the species volumes per unit channel width $D_i(x)=q{c}_{i0}(1-e^{-v_{i}x/q})$. The cumulative particle-volume fraction isFor both $v_i>0$, its limit at infinity is the incoming particle-volume fraction, whereas its limit as $x\downarrow 0$ is the local deposition fraction at the inlet. Local sorting and the cumulative inventory are different observables.

For an axisymmetric vertical mode of a shearing sheet in this Boussinesq approximation, the perturbations have no horizontal gradients, so background-shear advection vanishes. The linear Coriolis acceleration and perturbation advection of the background shear givewhile thermal conduction and incompressibility giveHere $\beta$ is the thermal diffusion rate of a Fourier mode, not the cooling exponent in Question 2. Since $k\ne 0$, incompressibility forces $u_z^{\prime }=0$, and the vertical momentum equation then forces $P^{\prime }=0$ for this nonzero-wavenumber amplitude.

The three remaining amplitudes satisfyA nontrivial solution requires the determinant to vanish, giving the convective overstability dispersion relationNo division by $s$ or $s+\beta$ was needed, so the thermal energy mode of a shearing sheet is retained. The notation $N^2$ denotes a signed buoyancy frequency squared and can be negative for an adverse stratification; it must not be restricted to the square of a real positive frequency.

Adopt the Boussinesq approximation, hydrostatic approximation, traditional approximation, and inviscid quasi-geostrophic approximation. Assume stable constant $N_0^2>0$, nonzero Coriolis parameter $f_0$, a beta plane, small Rossby number, $\beta L/|f_0|$ of that small order, and small displacement of the background stratification. Vertical advection of the buoyancy perturbation is then higher order. Define the quasi-geostrophic streamfunction byThe buoyancy and vertical vorticity equations at the retained order areDifferentiate the first in $z$. Since $J({\psi }_z,{\psi }_z)=0$, its material-derivative term becomes $f_0{D}_g{\psi }_{zz}$. Elimination of $w_z$ gives diabatically forced quasi-geostrophic potential vorticity:Thus it is the height gradient of heating, rather than heating alone, that produces potential vorticity at this order.

For the steady response linearized about rest, assume $\beta \ne 0$. The quasi-geostrophic potential-vorticity equation reduces to $\beta {\psi }_x=f_0{R}_z/N_0^2$. Choosing the horizontally uniform pressure gauge so that $\psi \to 0$ in the east gives the steady quasi-geostrophic response to localized heatingSetUsing the error function to integrate the Gaussian function givesThese velocities vanish as $x\to +\infty$. A depth-dependent but horizontally uniform addition to $\psi$ is dynamically irrelevant for the requested horizontal velocity. At $z=0$, streamlines are contours of $e^{-y^2/L^2}I(x)$. For $f_0,\beta ,R_0>0$, flow approaches the heating region from the west on its northern side, turns southward, and returns westward on its southern side. The contours are open, not closed gyres: $I$ tends to a nonzero constant in the west.

For $-H<z<0$, $R_z<0$ and the forcing decreases potential vorticity; steady balance requires $\beta v_g=f_0{R}_z/N_0^2<0$, giving southward flow. Below $z=-H$, $R_z>0$ and the meridional flow reverses. A constant heating rate independent of $z$ would have no such interior potential vorticity source.

For the nonlinear diagnostic quasi-geostrophic omega equation, write $\zeta ={\nabla }_h^2\psi$. Apply ${\nabla }_h^2$ to buoyancy evolution and $f_0{\partial }_z$ to vorticity evolution:Subtracting eliminates the pressure tendency and yieldsThe beta term is linear and must remain when the nonlinear terms are neglected. For the steady weak-forcing solution, $f_0\beta {\psi }_{xz}=f_0^2{R}_{zz}/N_0^2$, soThis also follows immediately from the steady linear buoyancy equation. It satisfies $w=0$ at $z=0$, decay at depth and horizontal infinity, and the linear quasi-geostrophic omega equation; with these homogeneous conditions the elliptic problem has no extra decaying homogeneous solution. The rigid-lid interpretation is an additional boundary approximation for this diagnostic response.

At $x=0$, the displayed $v_g$ and $w$ provide the requested meridional/vertical arrows. For positive heating there is upwelling concentrated near $y=0$, maximal in depth at $z=-H$, with northward flow below that depth and southward flow above. The vertical circulation tends to zero far from the forcing. A western zonal wake persists, but its meridional and vertical components vanish as $x\to -\infty$. Horizontal ageostrophic flow supplies the divergence associated with the upwelling; the projected $(v_g,w)$ slice is not itself a two-dimensional incompressible flow and need not have closed streamlines.

The steady calculation fixes an interior circulation, not its attainability from every possible initial and boundary state. For example, a rigid lid with initially uniform boundary buoyancy conserves that buoyancy at linear order because $R(0)=w(0)=0$. The formal steady profile instead generally has nonuniform $b^{\prime }(0)=f_0{\psi }_z(0)$. Such additional initial boundary data require a time-dependent adjustment and cannot simply be imposed on this particular steady solution. No boundary buoyancy data are supplied for the requested steady problem.

Let $p$ denote the fluid pressure perturbation divided by the reference mass density, and let $b=-g{\rho }^{\prime }/{\rho }_0$ be the buoyancy perturbation. The buoyancy frequency satisfies $N^2=-g{\hat{\rho }}_z/{\rho }_0$. Linearization requires small boundary slope $k|{\eta }_0|\ll 1$, small displacements compared with the vertical scales of the disturbance and background, and advection small compared with the oscillatory acceleration. For a propagating internal gravity wave this includes $|m{\eta }_0|\ll 1$; boundary slope alone is insufficient when $|m|\gg k$. The Boussinesq approximation also requires small relative mass density differences over the region of interest. These conditions must hold for the resulting disturbance, including any amplification by resonance.

The inviscid Linearized Boussinesq equations and linear kinematic boundary condition areWrite the velocity field as $(u,w)=\mathrm{Re}\{(U(z),W(z))e^{i(kx-\omega t)}\}$. Incompressibility and horizontal momentum balance giveThe vertical momentum balance therefore yieldsFor a real vertical wavenumber $m$, this gives the dispersion relationWhen $0<\omega <N$, define $\ell =k\sqrt{N^2/{\omega }^2-1}$. The radiation condition selects $m=-\ell$, because energy must leave the boundary upward. The boundary-forced internal gravity wave isIts wavevector is $\mathbf{K}=(k,m)$, its phase velocity is ${\mathbf{c}}_p=\omega \mathbf{K}/(k^2+m^2)$, and its group velocity isThus the phase velocity points down and right, while the group velocity points up and right. They are perpendicular. Internal-wave polarization makes the particle motion an oscillation along the group velocity direction, with zero first-order mean transport. The constant-phase lines of an internal gravity wave are also parallel to the group velocity. Their inclination $\vartheta$ above the horizontal satisfies $\sin \vartheta =\omega /N$.

When $\omega >N$, set $\kappa =k\sqrt{1-N^2/{\omega }^2}$. Boundedness at infinity selects the evanescent waveThe horizontal and vertical velocity components are in quadrature: fluid particles describe small ellipses, and the response decays over ${\kappa }^{-1}$. There is no upward time-averaged energy flux, because $P$ and $W$ are in quadrature. A real vertical group velocity is not defined for this evanescent wave. The pattern travels horizontally with phase velocity $\omega /k$.

For $N=0$ the response is the evanescent wave with $\kappa =k$; the ratios $\omega /N$ should not be used. At the cutoff $\omega =N>0$, the bounded harmonic solution has $W=-i\omega {\eta }_0$, $U=0$: it neither decays nor has nonzero upward group velocity. It is the limiting cutoff response, rather than a localized radiating disturbance. A uniform nonzero buoyancy frequency in an infinitely deep Boussinesq approximation is itself a local idealization of the background mass density.

The arrows distinguish the phase velocity, group velocity, and oscillatory particle motion of the internal gravity wave. The right panel shows the decay envelope and particle ellipses of the evanescent wave.

Use the shallow-water approximation, a hydrostatic vertically mixed layer, a deep motionless ambient, and no ambient fluid entrainment. Particle loss changes reduced gravity but changes total layer volume only at the discarded dilute-particle order. Likewise thermal expansion changes the equation of state while the leading Boussinesq approximation retains volume conservation. DefineDepth integration of mass conservation, horizontal momentum balance, particle transport, and the heating law givesHere the integrated excess hydrostatic pressure is $b{h}^2/2$. There is no particle source from the bed. Bed drag and the particle contribution to inertia are neglected at this order. The equivalent variable-buoyancy shallow water equations areThe term $h{b}_x/2$ is necessary when heating or particle loss makes the reduced gravity vary horizontally.

For $v=(h,u,\varphi ,\theta {)}^T$, the quasilinear system $v_t+A(v)v_x=S(v)$ hasIts characteristic polynomial factors asThus the characteristic curves have slopesOn the two repeated contact characteristic curves, the particle and heating laws are $D\varphi =-W_s\varphi /h$ and $D\theta =\beta \varphi$. Writing $c_g=\sqrt{bh}$ and $D_{\pm }={\partial }_t+(u\pm c_g){\partial }_x$, the gravity-wave compatibility relations areFor $h>0$ and $b>0$, gravity-wave right eigenvectors can be taken as $(1,\pm c_g/h,0,0{)}^T$. The contact eigenspace has dimension two: $\delta u=0$ and $b\delta h+(h/2)\delta b=0$, with independent $\delta \varphi ,\delta \theta$. Therefore $A$ has a full set of real eigenvectors.

The system is hyperbolic throughout strict static stability, but not strictly hyperbolic because the contact speed is repeated. At $b=0$ the eigenvalues coalesce and this diagonalizability is lost; when $b<0$ the gravity-wave speeds become complex, reflecting the loss of a statically stable lower layer.

It is important to fix the mass flux convention. With literal mass units and the corresponding density-weighted buoyancy flux, the Boussinesq approximation and circular top-hat plume model giveReplacing $\rho$ by ${\rho }_0$ in inertial fluxes is the Boussinesq approximation; the density difference is retained in the buoyancy flux. Define the kinematic plume fluxesIf the symbols $Q,\mathbf{M},F$ instead denote the customary kinematic fluxes, simply omit the factors ${\rho }_0$ in all conversions below. Keeping the two conventions consistent is essential for the jet length and dimensional prefactors.

The Batchelor entrainment hypothesis prescribes inward edge velocityThe constant entrainment coefficient relates ambient inflow to the local axial velocity. A centreline segment of arc length $ds$ entrains volume $2\pi b{u}_{e}ds$. Hence, setting $E=2\alpha \sqrt{\pi }$,The buoyancy flux is conserved because the homogeneous entrained ambient has zero density deficit, and mixing has no buoyancy source or sink. The imposed zero source mass flux with nonzero source momentum flux is a singular point-source idealization, not a finite-radius nozzle with zero emitted fluid. Indeed $V=m/q$ and the plume reduced gravity $f/q$ diverge as $q\to 0$ with nonzero source fluxes. The Boussinesq approximation cannot remain valid arbitrarily close to that point. The top-hat plume model is applied outside the unresolved source region.

Let $V=U-c$ and take $k>0$ without loss of generality. The Boussinesq approximation replaces density by ${\rho }_0$ in inertia and pressure acceleration, while retaining the density perturbation in buoyancy. Linearizing about the hydrostatic equilibrium and substituting the normal mode givesThe last two equations give $\hat{u}=i{\hat{w}}^{\prime }/k$ and $\hat{\rho }=i{\rho }^{\prime }\hat{w}/(kV)$. The horizontal momentum equation givesSubstituting in the vertical momentum equation yields the Taylor–Goldstein equation:The buoyancy frequency is real for stable density stratification. The impermeable walls impose $\hat{w}(\pm L)=0$.

Put $D={\partial }_t+U{\partial }_x$, write $p$ for perturbation pressure divided by ${\rho }_0$, and set $b=-g{\rho }^{\prime }/{\rho }_0$. The linearized Boussinesq approximation, incompressibility and material derivative giveTaking the curl eliminates pressure; differentiating the resulting vorticity equation in $x$ and using incompressibility gives $D{\nabla }^{2}w-U^{\prime \prime }{w}_x=b_{xx}$. A further application of $D$ therefore yieldsFor a stationary nonzero horizontal Fourier mode, $D=U{\partial }_x$, so cancellation of ${\partial }_x^2$ gives the stationary Taylor–Goldstein equationThis division requires $U\ne 0$ on the interval considered: a zero of $U$ is a critical level of an internal gravity wave. The mean profiles must be sufficiently smooth for the displayed derivatives, with background hydrostatic pressure and stable density stratification, $N^2>0$, for propagating internal gravity waves. The horizontally uniform component is excluded from the cancellation. For a horizontal wavenumber $k$, local vertical propagation additionally requires $N^2/U^2-U^{\prime \prime }/U>k^2$; this is distinct from the smoothness restrictions. If stability of the background against other disturbances is needed, the Miles–Howard theorem supplies the sufficient condition $N^2\ge (U^{\prime }{)}^2/4$, rather than a necessary condition for deriving the equation.

For a dilute particle suspension, take $W_s>0$ as the downward speed relative to the carrier fluid. The particle deposition flux through a horizontal absorbing bed is $W_s{\varphi }_b$, where ${\varphi }_b$ is the near-bed particle volume fraction. Without stirring, settling can create a particle-free upper layer and a descending concentration front, so uniform concentration throughout the remaining fluid is not automatic. Weak turbulence can maintain a nearly uniform suspended concentration while leaving the mean settling velocity approximately equal to the isolated-particle value; this justifies using $W_s\varphi$ in a mixed-layer balance. Strong stirring can produce resuspension and requires a separate erosion model.

For a bed inclined at angle $\vartheta$ to the horizontal, the normal settling speed is $W_s\cos \vartheta$, giving deposition flux $W_s{\varphi }_b\cos \vartheta$ per unit actual bed area. There is also downslope particle motion, and near-wall effects or resuspension can alter this simplest absorbing-boundary model. Integrating the normal projection over a sloping bed is equivalent to integrating $W_s{\varphi }_b$ over its horizontal projection.

The mixture mass density is $\rho ={\rho }_0+({\rho }_p-{\rho }_0)\varphi$, so its reduced gravity isThe Boussinesq approximation requires $({\rho }_p-{\rho }_0)\varphi /{\rho }_0\ll 1$, in addition to the dilution assumption.

Take inward fluid entrainment speed to be positive. The non-Boussinesq form of the Batchelor entrainment hypothesis iswhere $\alpha$ is the entrainment coefficient. Define physical, cross-section-integrated fluxes for the top-hat plume model byThus $Q$ is mass flux, $M$ is momentum flux, and $F$ is a density-weighted buoyancy flux. The kinematic buoyancy flux in the Boussinesq approximation is $F/{\rho }_0$. These definitions keep the factor $\pi$; conventions suppressing it simply rescale all three fluxes consistently.

For a homogeneous ambient, the sectional volume, mass and momentum balances areThe momentum source is the net upward buoyancy force; entrained ambient fluid initially supplies no vertical momentum. Subtracting the mass balance from ${\rho }_0$ times the volume balance gives conservation of the density deficit:Now $w=M/Q$, $\pi b^2\rho =Q^2/M$, and $\pi b^{2}g({\rho }_0-\rho )=QF/M$. Substitution produces the non-Boussinesq top-hat plume equationsFor reconstructing the fields without a Boussinesq approximation,The last expression uses the plume density in the acceleration denominator; it agrees with the ambient-density definition of reduced gravity to Boussinesq order.

Use the kinematic fluxes $q=Q/(\pi {\rho }_0)$, $m=M/(\pi {\rho }_0)$ and $f_0=F_0/(\pi {\rho }_0)$. In the steady Boussinesq approximation, the non-Boussinesq top-hat plume equations reduce toEliminating $z$ gives $m^{5/2}=5f_0{q}^2/(8\alpha )$, where the integration constant vanishes for a pure plume from a point source. Write $b=az$ and $w=K{z}^{-1/3}$. The two balances give $a=6\alpha /5$ and $K^3=3f_0/(4a^2)=25f_0/(48{\alpha }^2)$. The Boussinesq point-source plume is thereforeHere $q=a^{2}K{z}^{5/3}$ and $m=a^2{K}^2{z}^{4/3}$ tend to zero at the source, while $f_0$ remains positive. The plume Froude number is independent of height:The ideal point source is a far-field similarity idealization. Since $g^{\prime }\propto z^{-5/3}$, the Boussinesq approximation fails near $z=0$ and is valid only when $g^{\prime }/g\ll 1$. In terms of $K$, this requires $z\gg [4K^2/(3g){]}^{3/5}$; the density formula must not be extrapolated into its unphysical negative-density region.