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For a physical adiabatic ideal gas one has $\gamma >1$; the following algebra also holds for positive $\gamma \ne 1$. Compression work implies $p=\rho W^{\prime }(\rho )-W$, obtained by differentiating the total internal energy $VW(M/V)$ at fixed mass. With the stated pressure law this integrates toThe term proportional to mass is an arbitrary energy reference; set $C=0$. Mass flux is $\rho u$, momentum flux is $p+\rho u^2$, and dividing the energy flux by mass flux givesfor steady flow.

Work in the shock's rest frame. If the shock speed in the laboratory is $D$, upstream speed relative to it is $D$ and downstream speed is $D/r$, where $r={\rho }_1/{\rho }_0$. Conservation of momentum gives $p_1-p_0={\rho }_0{D}^2(1-1/r)$. Conservation of specific total enthalpy givesEliminating $D^2$ and using $p_1/p_0=1+\beta$ yieldsIts weak-shock expansion is $1+\beta /\gamma -(\gamma -1){\beta }^2/(2{\gamma }^2)+O({\beta }^3)$, exactly the expansion of $(1+\beta {)}^{1/\gamma }$ through second order. Thus the adiabatic pressure-density relation agrees to order ${\beta }^2$; entropy production first affects this comparison at higher order. The same isentropic constant is not exactly preserved across a finite shock. The wording allows $\gamma =1$, but the printed enthalpy formula is then undefined. The compression calculation instead gives $W=(p_0/{\rho }_0)\rho \log (\rho /{\rho }_0)+C\rho$, and the specific enthalpy is $(p_0/{\rho }_0)[1+\log (\rho /{\rho }_0)]+C$. Thus the requested formula containing $1/(\gamma -1)$ requires $\gamma \ne 1$; its physical ideal-gas application uses $\gamma >1$.

On one poloidal flux surface, the field-line angular velocity $\omega (\psi )$ is constant. In a frame rotating at that rate the relative velocity is $\mathbf{v}=\mathbf{u}-R\omega {\mathbf{e}}_{\varphi }=k\mathbf{B}/\rho$, parallel to the magnetic field. The Lorentz force is perpendicular to this motion, and the Coriolis force also does no work. Centrifugal acceleration comes from the effective potential $-R^2{\omega }^2/2$. Since entropy is constant along the surface, pressure work integrates to specific enthalpy $w$, with $dw=dp/\rho$ along the flow.

Therefore kinetic energy relative to the rotating poloidal flux surface, enthalpy and effective gravitational-plus-centrifugal potential have a constant sum:This is a corotating energy integral, not the laboratory energy alone: Maxwell stress tensor stresses can transfer energy and angular momentum in the laboratory frame. In fact the expression equals $|\mathbf{u}{|}^2/2+w+\Phi -\omega R{u}_{\varphi }$, namely the total magnetohydrodynamic Bernoulli invariant minus $\omega \ell$. Differential rotation between surfaces does not invalidate the argument, because the chosen angular velocity is fixed along the particular surface being followed.

Let $e$ be the specific internal energy, $s$ the specific entropy, $T$ the absolute temperature, and $w=e+p/\rho$ the specific enthalpy. The first law of thermodynamics for a simple fluid of fixed composition is $de=Tds-pd(1/\rho )$, soThe unmagnetized Euler equations for an inviscid fluid are ${\partial }_t\mathbf{u}+(\mathbf{u}\cdot \nabla )\mathbf{u}=-\nabla \Phi -\nabla p/\rho$. Define the vorticity $\mathbf{\omega }=\nabla \times \mathbf{u}$ and use the vorticity cross-product identity $(\mathbf{u}\cdot \nabla )\mathbf{u}=\mathbf{\omega }\times \mathbf{u}+\nabla (u^2/2)$. Substitution gives the Crocco form of the unsteady Euler equation:Adiabaticity means $Ds/Dt=0$ for smooth flow without dissipative heating. It does not require $s$ to be uniform between different fluid elements, so the entropy-gradient term generally remains. For a perfect gas with constant adiabatic exponent $\gamma$, one may write $w=\gamma p/[(\gamma -1)\rho ]$, but the thermodynamic derivation does not require that particular equation of state.

Use a one-sided top-hat plume model with Boussinesq approximation and neglect the solid/liquid density difference in the interfacial volume source. Define the positive ablation speed $U=-\dot{a}$, volume flux per unit wall span $Q=bw$, and kinematic momentum flux $J=b{w}^2$. Let ambient entrainment per height be $E=\epsilon w$, with entrainment coefficient $\epsilon$, and take $\rho ={\rho }_{\infty }[1+{\beta }_C(C-C_{\infty })]$ for the salinity contribution to density. Conservation of volume and vertical momentum, with negligible wall stress and negligible vertical momentum of entering fluid, givesOnly the ocean-facing edge entrains. Using two edges would change the entrainment coefficient in these formulas.

To derive the heat balance, choose liquid specific enthalpy $h_l=c_{p}T+L$ and solid specific enthalpy $h_s=c_p{T}_s$. A horizontal slab receives ambient volume $Edz$ with enthalpy $h_{\infty }=c_p{T}_{\infty }+L$, and solid mass-equivalent volume $Udz$ with enthalpy $h_s$. ThusSubtract $(c_p{T}_{\infty }+L)Q^{\prime }$ and divide by $c_p$:Similarly, the incoming salt flux is $E{C}_{\infty }+U{C}_s$, so $(QC{)}^{\prime }=E{C}_{\infty }+U{C}_s$. Subtracting $C_{\infty }{Q}^{\prime }$ provesBoth anomalies are negative for fresh, cold melt entering the ocean. These balances account for the heat needed to melt the entering ice, rather than treating it as already-liquid water at $T_s$.

Now use $L/c_p\gg T_{\infty }-T_s$, the large latent-to-sensible heat ratio convention of the paper. Put $\Delta C=C_{\infty }-C_s>0$. The stated small plume-deficit limits allow us to replace plume $T,C$ by their ambient values in the diffusive interfacial fluxes. The Stefan condition and interfacial salt balance are thenFor the compositional Péclet number $P\equiv P{e}_c=U\delta /D$, solve the second equation and impose the liquidus:Substitution in the heat balance gives the diffusive glacier-ablation interface lawHere $\Gamma$ is the liquidus slope, unrelated to the surface-energy parameter of the previous question. If $A_T=LD/(c_p\kappa )$, $\Theta =T_{\infty }-T_m$ and $B_T=\Gamma \Delta C$, the physical root isIt is positive when $\Theta +B_T>0$, exactly the condition that the ocean is warmer than its own liquidus. Since $\delta$ and ambient properties are fixed, $U$ is constant with height in this approximation.

For the melting-driven saline wall plume, measure height $Z=z-z_0$ from a virtual origin. The integrated anomaly balances areIn the specified small-deficit regime the meltwater volume is negligible compared with entrainment. Indeed, the solution below gives $(C_{\infty }-C)/\Delta C=4U/(3\epsilon w)$, so a small compositional plume deficit implies $U\ll \epsilon w$. Thus retain $Q^{\prime }=\epsilon w$ for the leading power laws, although $U$ remains essential in the salt and heat sources. Try $b=BZ$, $w=W{Z}^q$. The salt deficit is proportional to $Z^{-q}$, and the momentum balance has powers $Z^{2q}=Z^{1-q}$, hence $q=1/3$. The volume balance gives $(4/3)BW=\epsilon W$, and the momentum coefficient gives $(5/3)B{W}^2=g{\beta }_{C}U\Delta C/W$. ThereforeThese are entrainment-dominated power laws, not an exact solution with the $U$ term retained in the volume equation. They fail close to the virtual origin, where the formal temperature and salinity deficits diverge. Their buoyancy source is distributed melting along the wall, unlike a line plume with a constant buoyancy flux from a localized source.

The profiles rise from $T_i,C_i$ across the laminar boundary layer to the almost-ambient plume values $T,C$, then approach $T_{\infty },C_{\infty }$ across the entraining edge. The sketch uses the linear-gradient approximation implicit in the prescribed diffusive fluxes; it is not an exact normal advection–diffusion profile at large $P$.

For $P\ll 1$,The last formula retains both first-order terms and is appropriate near the no-ablation threshold. Salt diffusion keeps the interface saline, and freezing-point depression allows ice dissolution in saline water even when $T_{\infty }<T_m$. Latent heat is still needed, but the thermodynamic permission to ablate comes from the depressed liquidus; small $P$ does not mean a negligible energy cost.

For $P\gg 1$, $C_i-C_s\sim \Delta C/P$ and $T_i\sim T_m-B_T/P$. Rapid production of fresh melt dilutes the interface, so its temperature approaches the pure-ice melting point. In the usual thermally driven regime $\Theta >0$ and $\Theta \gg B_T/P$,This is predominantly thermal melting. More generally the retained large-$P$ balance is $A_{T}P\simeq \Theta +B_T/P$. For example, $\Theta =0$ gives $P\sim \sqrt{B_T/A_T}$, which can be large while the driving term is still saline freezing-point depression. Thus the melting/dissolution interpretation must also consider the ocean's temperature relative to $T_m$, not the compositional Péclet number alone.

Let $u=u_r<0$ for accretion. The steady spherical continuity equation and radial Euler equations for an inviscid fluid giveFor a barotropic fluid, $dp/dr=c_s^{2}d\rho /dr$. Dividing continuity equation by $r^2\rho u$ gives ${\rho }^{\prime }/\rho =-2/r-u^{\prime }/u$. Eliminating this derivative yieldsAt a smooth transonic critical point, the coefficient of $du/dr$ vanishes. A finite derivative then requires the right side to vanish simultaneously:These are the sonic and regularity conditions. Otherwise the derivative is singular and the proposed smooth transonic passage fails. The local slope must also be a real root compatible with the desired branch. For example, defining $C=\rho d(c_s^2)/d\rho$ at the point and differentiating both sides givesThis explains why simultaneous vanishing is a necessary regularity condition, rather than an automatic proof that every potential admits a critical crossing.

Define the barotropic enthalpy function by $dh=dp/\rho$. The momentum equation integrates to the Bernoulli equationChanging the reference mass density in $h$ changes only the constant $B$. For an isothermal closure this logarithmic function is a barotropic barotropic pressure potential; it should not be confused with the constant thermodynamic specific enthalpy of an ideal gas held at fixed temperature.

For isothermal transonic accretion in the Paczyński-Wiita potential, assume $M>0$ and $c_s>0$, and define $r_B=GM/(2c_s^2)$ and $\beta =16c_s^2/c^2$. The critical equation iswhose two roots are $r_G+\frac{1}{2}{r}_B(1\pm \sqrt{1+\beta })$. Only the plus root lies outside $r_G$. ThereforeFor this isothermal case $C=0$, and the slope relation reduces toThe positive slope is the inward-accretion solution connecting a small inward speed at large radius to the supersonic inward branch. Thus the exterior critical point is a genuine nondegenerate transonic point.

Choose the reference mass density as ${\rho }_0$, so $h=c_s^2\log (\rho /{\rho }_0)$ and $B=0$ from the conditions at infinity. At $u_c=-c_s$,Substitution into $\dot{M}=4\pi r_c^2{\rho }_c{c}_s$ givesThis is the rate selected by the smooth transonic accretion solution. Arbitrary static or subsonic formal solutions are not assigned this rate merely by specifying conditions at infinity. As $r_G/r_B\to 0$, the result tends to the isothermal Bondi accretion rate $\pi e^{3/2}{\rho }_0(GM{)}^2/c_s^3$, a useful normalization check.

Let $\Psi$ be the gravitational potential, with ${\nabla }^2\Psi =4\pi G\rho$, and let ${\varpi }^2=x^2+y^2$ measure distance from the rotation axis. Uniform solid-body rotation contributes the centrifugal potential $-{\Omega }^2{\varpi }^2/2$. In the corotating frame, static force balance therefore gives the uniformly rotating stellar hydrostatic equilibrium equationsThe final coefficient follows because ${\nabla }^2(x^2+y^2)=4$.

A tangent displacement along a regular connected equipotential surface has zero dot product with $\nabla \varphi$, and hence with $\nabla P$. Thus $P=P(\varphi )$ locally. Taking the curl of the hydrostatic equilibrium equation gives $\nabla \rho \times \nabla \varphi =0$, so $\rho =\rho (\varphi )$ on such surfaces as well. The Poisson equation then shows that ${\nabla }^2\varphi$ is constant on each equipotential surface.

The magnitude $g=|\nabla \varphi |$ need not be constant. For instance $\varphi =A(x^2+y^2)+B{z}^2$, with positive $A\ne B$, has constant Laplacian but different pole and equator gradient magnitudes on an ellipsoidal equipotential. In a rotating star, flattening and centrifugal acceleration likewise make the effective gravity latitude dependent.

With uniform composition, an equation of state determines $T$ from $P$ and $\rho$; for the supplied ideal gas plus radiation pressure law its right side increases monotonically with $T$. Hence $T=T(\varphi )$, and the radiative conductivity $\chi (\rho ,T)$ is also constant on an equipotential surface. Put $f(\varphi )=-\chi dT/d\varphi$. Then $\mathbf{F}=f(\varphi )\nabla \varphi$. On the stellar surface, below mass shedding, the outward normal is $\nabla \varphi /g$, soThis is the radiative gravity-darkening law. The effective temperature measures emergent flux; it is not the same quantity as the interior material temperature that was constant on an equipotential.

A rapidly rotating star is oblate, with reduced effective gravity and cooler effective temperature around the equator. The poles are hottest. The original sketch uses a Roche model of a uniformly rotating star only to illustrate the shape and surface-flux variation:

The obstruction to purely local radiative equilibrium is visible by taking the divergence:On an equipotential surface, $\rho ϵ$, $f$, $f^{\prime }$ and the second term are constant, but $g^2$ generally varies. Unless $f^{\prime }=0$ or another special circumstance removes the mismatch, $\nabla \cdot \mathbf{F}=\rho ϵ$ cannot hold at every latitude. This is the radiative-equilibrium obstruction in a rotating barotropic star; heat advection by stellar meridional circulation resolves the missing energy transport in the leading hydrostatic treatment.

For the steady circulation, mass conservation gives $\nabla \cdot (\rho \mathbf{v})=0$. The specific enthalpy identity and hydrostatic equilibrium giveConsequentlyBoth $h$ and $\varphi$ are constant on the bounding equipotential surface. By the divergence theorem and steady continuity equation,Integrating the given thermal energy equation therefore gives equipotential luminosity conservation with stellar circulation:Individual circulation streams can cross the surface; only their net steady mass flux has to vanish.

Since $f$ is constant on that surface, integrating the Poisson equation givesThus $dT/d\varphi =-L/[\chi (4\pi Gm-2{\Omega }^{2}V)]$, while $dP/d\varphi =-\rho$. Dividing these derivatives and inserting $\chi =4ac{T}^3/(3\kappa \rho )$ yields the radiative gradient in a uniformly rotating star:The bracket must be positive for the regular outward-gravity configuration being used.

For an interior equipotential, both mass factors must be the enclosed mass $m$. The printed final bracket uses $M$, but its derivation gives $m$; at the outer surface $m=M$ and the printed bracket is recovered. For a direct check, a uniform-density model has $m=\rho V$, so the true correction ${\Omega }^{2}V/(2\pi Gm)={\Omega }^2/(2\pi G\rho )$ is independent of the chosen enclosed volume. Replacing $m$ by a fixed total mass would incorrectly make it volume dependent.

Integrating each conservation law across a vanishingly thin interval around the stationary shock wave leaves equal conservation law fluxes on its two sides. HenceThese are the Rankine-Hugoniot conditions for a perfect gas. Signed velocities may both be negative when the material travels from positive to negative $z$; no sign change is needed in the conservation law fluxes.

Put $r={\rho }_2/{\rho }_1=u_1/u_2$, $P=p_2/p_1$ and $m=M_1^2={\rho }_1{u}_1^2/(\gamma p_1)$. Momentum conservation givesDivide the energy condition by the nonzero mass flux. Equality of kinetic energy plus specific enthalpy givesSubstituting $P$ and multiplying by $2(\gamma -1)r^2/\gamma$ yieldsThe factor $r-1$ is the continuous, no-shock solution. On the nontrivial normal shock wave branch,An admissible compressive gas shock wave has $M_1>1$, so $r>1$ and $P>1$. The algebraic jump equations alone also allow a reversed expansive discontinuity; the entropy production in a perfect-gas shock excludes that branch. At $M_1=1$ the nontrivial branch joins the continuous solution.

For steady spherical polytropic flow, write $u=u_r$ and use the polytropic equation of state $p=K{\rho }^{\gamma }$, with $K>0$. Mass conservation and radial Euler momentum equation areThe mass equation gives ${\rho }^{\prime }/\rho =-2/r-u^{\prime }/u$. Substitute it into momentum balance to obtainAt a sonic point the derivative coefficient vanishes. A smooth finite-slope solution must make the numerator vanish there too:Finally $dp/\rho =d[\gamma K{\rho }^{\gamma -1}/(\gamma -1)]$. Integrating momentum gives the Bernoulli equationThis is kinetic energy plus specific enthalpy plus gravitational potential per unit mass. The polytropic stellar wind selects a particular transonic branch of these equations.

For cylindrical rotation $u_{\varphi }=R\Omega (R)$, the equilibrium Euler momentum equation is $-R{\Omega }^2{\mathbf{e}}_R=-\nabla p/\rho -\nabla \Phi$. HenceThe integral is the centrifugal potential of cylindrical rotation. On a regular barotropic fluid branch define the specific enthalpy $h(\rho )$ by $dh=dp/\rho$. ThenFor a regular equation of state with $dp/d\rho >0$, $h$ is invertible, so $\rho$ and $p=p(\rho )$ depend only on $\Psi$ within a connected equilibrium region. In particular $dp/d\Psi =-\rho$. This local conclusion uses an invertible barotropic branch; distinct disconnected fluid regions can have different integration constants.

The circular Keplerian orbit has speed $v_K=(GM/r{)}^{1/2}$. Equating it with the heated gas sound speed givesThis is the photoevaporative gravitational radius, where thermal and orbital binding energies have the same order of magnitude. A circular orbit has specific mechanical energy $-GM/(2r)$. Heating adds thermal energy and, in a fluid outflow, available specific enthalpy of order $c^2$. For example, if $c$ is the adiabatic sound speed, an ordinary ideal gas has enthalpy $c^2/(\gamma -1)$; for $\gamma =5/3$ this is $3c^2/2$. At $r\ge r_g$, this more than compensates the circular-orbit binding energy. Equivalently the hot hydrostatic scale height satisfies $H_{\mathrm{hot}}/r\sim c/v_K\gtrsim 1$, so a thin bound surface layer cannot be maintained. With continued irradiation, the gas can expand into a thermal wind: photoevaporation removes disk material.

The condition is a thermal binding scale, rather than an assertion that the sound speed equals the ballistic escape speed, which is $\sqrt{2}{v}_K$. Detailed wind launching can change the numerical critical radius by factors of order unity. Using exactly the supplied numerical estimates, $c=10^6\mathrm{cm}{\mathrm{s}}^{-1}$, and therefore

Write the radial velocity as $u(r)$, allowing $u<0$ for accretion flow, and let $c^2=\gamma p/\rho$ be the squared adiabatic sound speed. The steady continuity equation, radial Euler equations for an inviscid fluid and entropy advection equation giveOn a nonzero smooth flow branch, the polytropic equation of state is $p=K{\rho }^{\gamma }$ with constant $K$. Thus the two useful first integrals for spherically symmetric adiabatic flow areHere $\mathcal{J}$ is the signed mass flux and $\mathcal{B}$ is the Bernoulli function. The specific enthalpy is $c^2/(\gamma -1)$; the prescribed Newtonian gravitational potential has no contribution from the gas's own gravity. Eliminating the density derivative using mass conservation gives the differential formThis sonic point equation displays both the singular coefficient and the numerator that must vanish for a smooth transonic branch.

Use the printed stellar gas-pressure fraction, $\beta =P_g/P$, with $P=P_g+P_{\mathrm{rad}}$, and keep composition fixed. For a monatomic perfect gas plus equilibrium blackbody radiation, the specific heats of a monatomic gas-radiation mixture follow from its specific internal energy and specific enthalpy:At fixed total pressure, differentiating the equation of state givesIn particular, $\beta$ is not held constant during that derivative. The specific heat capacity at constant pressure is $c_P=(\partial h/\partial T{)}_P$, soSince $a{T}^3/(3\rho )=\mathcal{R}(1-\beta )/\beta$, this simplifies toThe pure-gas limit is $5\mathcal{R}/2$. For a gas with unspecified molecular degrees of freedom, replace $3\mathcal{R}/2$ in $u$ by its gas heat capacity $c_{V,g}$: then $c_P=c_{V,g}+\mathcal{R}+4\mathcal{R}(1-\beta )(4+\beta )/{\beta }^2$. The boxed formula uses the conventional monatomic stellar-gas interpretation.

Let $e$ be specific internal energy, $Q$ the heat supplied per unit volume per unit time, and $E=\rho (e+|\mathbf{u}{|}^2/2)$. For an inviscid perfect gas in a prescribed Newtonian gravitational potential $\Phi$, the fluid total-energy equation isThere is no gravitational term if no external body is present. For a time-independent $\Phi$, adding potential energy gives the equivalent conservative form ${\partial }_t(E+\rho \Phi )+\nabla \cdot [(E+p+\rho \Phi )\mathbf{u}]=Q$. Subtracting the kinetic-energy equation and using mass conservation givesFor a perfect gas with constant specific-heat ratio $\gamma$, $e=p/[(\gamma -1)\rho ]$. The isothermal equation of state therefore makes $e=c_s^2/(\gamma -1)$ constant. Hence the required heat supply and heat loss areCompression requires cooling; expansion requires heating. The isothermal sound speed $c_s$ differs from the adiabatic sound speed $\sqrt{\gamma }{c}_s$.

Define the isothermal Mach numbers by ${\mathcal{M}}_i=u_i/c_s$. Orient the normal to a stationary isothermal shock along the flow, so the mass flux $j={\rho }_1{u}_1={\rho }_2{u}_2>0$. Tangential velocity is continuous for this planar hydrodynamic shock. Normal momentum conservation givesFor a genuine discontinuity $u_1\ne u_2$, factorization gives $u_1{u}_2=c_s^2$. Thus the isothermal shock jump relations areAcross a thin shock, gravitational potential and tangential kinetic energy are unchanged. The specific enthalpy $h=\gamma c_s^2/(\gamma -1)$ is the same on both sides, so the energy removed per unit area per unit time isPositive net cooling requires $u_1>u_2$, equivalently ${\rho }_2>{\rho }_1$ and ${\mathcal{M}}_1>1>{\mathcal{M}}_2$. Cooling across an isothermal shock permits only compression shocks; the formal expansion discontinuity requires energy supplied by the surroundings. The continuous state $u_1=u_2$ is not a shock.

For steady radial flow, mass conservation gives $r^2\rho u_r=\text{constant}$. Differentiating it and eliminating the mass density gradient from radial momentum balance givesA regular sonic point has $|u_r|=c_s$, so the right-hand side must vanish there as well. ThereforeThis is the Bondi sonic point for accretion; the same critical radius appears in an isothermal Parker wind. A general steady solution need not pass through a sonic point, but a smooth transonic one must obey both critical conditions.

Put $x=r/r_s$ and $\mathcal{M}=|u_r|/c_s$ on a branch whose flow direction is fixed. The radial equation becomesIntegrating yields the integrated isothermal Bondi flow relation:For a transonic branch, $x=\mathcal{M}=1$ gives $C=-3/2$. Expansion about the critical point gives $(\mathcal{M}-1{)}^2=(x-1{)}^2+O((x-1{)}^3)$, with slope $-1$ for inflow supplied by gas at rest at infinity and slope $+1$ for a transonic outflow.

For Isothermal Bondi accretion, integrate radial momentum balance once more, using $u_r\to 0$ and $\rho \to {\rho }_{\infty }$ at infinity:At the Bondi sonic point, $GM/r_s=2c_s^2$ and $u_r^2=c_s^2$, so ${\rho }_s=e^{3/2}{\rho }_{\infty }$. The positive inward isothermal Bondi accretion rate is consequentlyThe critical mass density and rate are fixed by the regular transonic solution and the specified reservoir at infinity; an arbitrary subsonic solution does not share this accretion rate.

For the polytropic equation of state $p=K{\rho }^{1+1/n}$, take $K>0$ and $n>0$ for the physical model. The specific enthalpy relative to vacuum isIts gradient obeys $\nabla Q={\rho }^{-1}\nabla p$, so the Euler equations becomeFor the material derivative $D_t={\partial }_t+\mathbf{u}\cdot \nabla$, $D_{t}Q=(Q/n\rho )D_t\rho$. The continuity equation thus becomesThe choice $Q=0$ at zero mass density fixes the additive convention in this equation. Although the momentum equation alone is unchanged by adding a constant to $Q$, the polytropic relation and enthalpy evolution must use the same vacuum reference.

Take $z$ positive downwards from the roof. Write $T_0=T_m-m{C}_0$, $C_B=(T_m-T_B)/m$, $\Delta T=T_0-T_B=m(C_B-C_0)>0$ and $\Delta C=C_B-C_0$. The cooled magma follows the decreasing liquidus from $(C_0,T_0)$ towards $(C_B,T_B)$; the initially superheated liquid starts at $(C_0,T_{\infty })$ above that liquidus. The solid has zero solute concentration. At each temperature in the mushy layer, a horizontal tie line joins the pure solid to the residual liquid on the liquidus. The lever rule gives the liquid fraction from the bulk concentration. Since $T_B>T_E$, this path stops before the eutectic temperature and does not cross the eutectic composition.