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Heating increases the monomer fraction through $\mathrm{A}{\mathrm{l}}_2\mathrm{C}{\mathrm{l}}_6(\mathrm{g})\rightleftharpoons 2AlC{\mathrm{l}}_3(\mathrm{g})$. Dissociation breaks the bridge interactions and gains translational entropy by making two gas molecules from one. Its enthalpy cost becomes easier to offset at higher temperature. The qualitative statement concerns the molecular gas-phase equilibrium at the stated conditions, rather than every solid or liquid phase of aluminium chloride.

Excess ethanol drives a reversible ester exchange mainly through the actual mixture composition, not through a large new bond-energy advantage. The net reaction exchanges one ester and one alcohol for a closely related ester and alcohol. It preserves the main bond types and has no obvious large exothermicity. Solvation and ion pairing can contribute to its enthalpy, but there is no basis here for asserting a large negative standard reaction enthalpy.

There are two molecules on each side, so an increase in particle count is not the entropy explanation. Instead useWith ethanol as the solvent and little methanol initially, the reaction quotient is small; product formation and mixing lower the Gibbs free energy until the exchange equilibrium is reached. ethanol can be treated approximately as having fixed activity near its pure-solvent value, rather than assigned an arbitrarily enormous activity, while methanol remains dilute. This composition-dependent chemical-potential contribution is the appropriate entropic/mixing argument. It explains forward conversion under these conditions without claiming that ${\Delta }_r{S}^{\circ }$ must be positive, or that complete conversion follows without an equilibrium constant or removal of methanol.

Let $G$ be Gibbs free energy, $H$ enthalpy and $T$ thermodynamic temperature for a fixed amount and composition of material, with the derivative taken at constant pressure. Apply the Gibbs–Helmholtz equation to the standard molar reaction quantities, obtained by summing product minus reactant contributions with their stoichiometric coefficients:Define $K(T)$ as the dimensionless thermodynamic equilibrium constant expressed using activities in specified standard states. Since ${\Delta }_r{G}^{\circ }=-RT\ln K$, where $R$ is the molar gas constant,If the reaction enthalpy is constant over the interval, integration givesThis is the requested Van't Hoff equation. For an exothermic reaction, increasing temperature decreases $K$. Although the name “isochore” is often used in this context, the displayed enthalpy form follows from standard Gibbs energies and the stated constant-pressure Gibbs–Helmholtz identity; the analogous ideal-gas concentration form involves internal energy when the temperature dependence of standard concentration is retained.

The two remaining partial pressures sum to $200-160=40$ bar. Their ratio therefore gives $p_{{\mathrm{N}}_2}=10$ bar and $p_{\mathrm{N}{\mathrm{H}}_3}=30$ bar. With the ideal-gas activity approximation $a_i=p_i/p^{\circ }$ and $p^{\circ }=1$ bar,The equilibrium constant is dimensionless: pressures enter as ratios to the standard pressure. Blindly substituting pressures in pascals into an unnormalized expression would give a different, incorrect thermodynamic $K$.

The standard reaction enthalpy is $2(-45.90)=-91.80\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ per mole of reaction as written, because the elements are in their reference states. ThusThis calculation uses the supplied constant-enthalpy simplification. At 200 bar, an accurate thermodynamic treatment would use chemical fugacities rather than ideal partial pressures; no chemical fugacity coefficients are supplied. The iron catalyst changes the approach to equilibrium, not its thermodynamic value.

Using bonds broken minus bonds formed, the enthalpies per mole of elementary events areUnder the steady-state approximation in chemical kinetics, initiation and termination have equal event rates, so their enthalpy contributions cancel. The two propagation steps also have equal flux, and their combined enthalpy is $5-189=\boxed{-184\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}}$ per overall reaction producing two moles of HCl. Step 3 dominates the heat release; the small endothermic contribution of step 2 only slightly offsets it.

An insulated vessel cannot remove this released energy as heat, so its temperature rises as reaction proceeds. In a rigid vessel the precise energy balance uses reaction internal energy and constant-volume heat capacities, whereas a constant-pressure adiabatic balance uses enthalpy; the strong exothermic conclusion is the same. Warming changes the rate constants, and activated initiation and propagation can accelerate, producing positive thermal feedback and potentially a rapid temperature rise. A numerical final temperature or a claim of an inevitable explosion is not justified without composition, heat capacities and heat-loss information. The initially assumed 298-K rate constants cannot remain fixed throughout an appreciable adiabatic temperature rise.

The unusually stable divalent configurations Eu(II) $4f^7$ and Yb(II) $4f^{14}$ make their third ionization particularly costly. Forming Eu(III), $4f^6$, or Yb(III), $4f^{13}$ removes an electron from a half-filled or filled f shell. The positive ionization-energy contribution therefore increases relative to the smoother background across the series.

Their trivalent ionic radii and hydration enthalpies do not supply a matching abrupt extra stabilization. Differences in atomization and the divalent character of elemental Eu and Yb partly compensate the ionization anomaly but do not erase the net less favourable trivalent aqueous formation enthalpy. This explanation concerns formation of the $+3$ aqueous ions, not a claim that $\mathrm{E}{\mathrm{u}}^{3+}$ or $\mathrm{Y}{\mathrm{b}}^{3+}$ is intrinsically the most stable isolated f configuration.

The large positive sum of ionization energies and large negative hydration enthalpy dominate, with a smaller but significant positive atomization term. Ionization depends on effective nuclear attraction, orbital energies and the stability of successive gas-ion electron configurations; removing the third electron may require disruption of a particularly stable divalent f configuration. Atomization depends on the metal's cohesive bonding and valence participation, and need not vary perfectly smoothly.

Hydration becomes more favourable for higher charge density. All the compared ions have charge $+3$, so smaller effective ionic radius generally gives more negative hydration enthalpy. Incomplete shielding by the 4f shell produces the lanthanide contraction and strengthens ion–water attraction along the series. A Born-type electrostatic estimate has a stabilization scale proportional to $-z^2/r$, but hydration-shell structure, water polarization and temperature dependence prevent treating that crude scale as an exact enthalpy formula. The hydration trend is smoother than the sharp configuration-dependent ionization anomalies. The measured formation enthalpy reflects cancellation of large terms, so comparison of just one term can give a misleading prediction.

At fixed particle number and pressure, the joint isothermal-isobaric ensemble weight for a microstate of energy $E$ and volume $V$ is $e^{-\beta (E+PV)}$. Write its enthalpy as $H=E+PV$. Differentiating the isothermal-isobaric partition function with a temperature-independent volume measure givesAfter division by $Q_P$ this is the normalized ensemble average, soThe differentiation assumes that the microscopic Hamiltonian and the integration measure have no additional explicit temperature dependence. An extra temperature-dependent normalization factor must be differentiated separately.

Differentiate the normalized average, including the derivative of its denominator. The enthalpy fluctuation formula follows fromSince $d\beta /dT=-1/(k_B{T}^2)$,Thus a larger equilibrium spread of enthalpy corresponds to a larger response to heating at fixed pressure. The variance is nonnegative, so this finite-system equilibrium description has $C_P\ge 0$. The relation connects a response coefficient to fluctuations of the appropriate exchanged quantity; at fixed volume the corresponding quantity is internal energy.

Away from singular regimes, assume enthalpy and heat capacity at constant pressure are extensive: $⟨H⟩=Nh$ and $C_P=N{c}_P$, with nonzero intensive $h$. The enthalpy fluctuation formula then givesAbsolute fluctuations grow as $\sqrt{N}$, while the mean grows as $N$. In the thermodynamic limit, enthalpy per particle therefore concentrates around its mean; it is relative, rather than absolute, fluctuations that vanish. This is the usual relative fluctuations of an extensive observable scaling. It assumes ordinary extensive heat capacity and a nonzero mean with a stated energy reference; a critical divergence or an artificially zero mean requires separate analysis.

Take a fixed amount of dry air, with no phase change, moving quasistatically so that its pressure matches the surrounding atmosphere. From the first law of thermodynamics and $H=U+pV$,For adiabatic motion $dq=0$. Treating the parcel as an ideal gas of fixed composition makes its enthalpy depend only on temperature, so for $n$ moles $dH=n{C}_{p}dT$ even though the pressure changes. Hence $n{C}_{p}dT=Vdp$. The environment is in hydrostatic equilibrium, $dp/dz=-\rho g$, and the parcel is taken to have the ambient density at the displacement point. Using $\rho =nM/V$ givesHere $C_p$ is molar and $c_p=C_p/M$ is mass-specific heat capacity. Positive $z$ points upward; the commonly quoted positive dry adiabatic lapse rate is ${\Gamma }_d=-dT/dz$. Approximate constancy of $g$ and heat capacity gives a constant gradient. Condensation, external heating, entrainment or composition changes invalidate this dry-parcel relation.

galactic feedback couples star formation and black-hole accretion back to the gas supply. Stellar radiation, winds, supernovae and cosmic rays inject energy and momentum; active-galactic-nucleus feedback adds radiation, jets and winds powered by accretion. Heating, turbulence or gas ejection can suppress later star formation, while compression can sometimes promote it. Negative feedback permits self-regulation rather than unrestricted gas consumption.

Define the mass-loading factor $\eta ={\dot{M}}_{\mathrm{out}}/{\dot{M}}_{\ast }$. If the launch speed scales with the escape speed and the injected energy per stellar mass is fixed, conservation of energy suggests $\eta \propto v_{\mathrm{esc}}^{-2}$; fixed injected momentum instead suggests $\eta \propto v_{\mathrm{esc}}^{-1}$. Coupling efficiencies, radiative losses and recycling change these ideal scalings. Thermal energy injected into dense gas can radiate away before driving an outflow, so an energy budget alone does not demonstrate effective feedback.

Radio-filled X-ray cavities provide a mechanical diagnostic. For a cavity containing a relativistic fluid with $\gamma =4/3$, its enthalpy is $\gamma PV/(\gamma -1)\simeq 4PV$; dividing by an estimated age gives a mechanical-power estimate. Chandra X-ray Observatory observations of the Perseus cluster revealed fronts and ripples associated with radio bubbles, interpreted as weak shocks and pressure waves that can transport jet energy into the cooling atmosphere.^[ref]

The test requires ambient density and temperature estimates, radio/X-ray spatial comparison, projected cavity volumes and an age model. Brightness edges must be distinguished from absorption and cool rims; wave energy is not automatically thermalized locally. These controls determine whether inferred mechanical heating can actually offset radiative cooling.

For a monatomic ideal gas, write the thermal energy density as $u=3n{k}_{B}T/2$. At low optical depth, radiative loss uses an astrophysical cooling function, often written $\mathcal{C}=n_e{n}_H\Lambda (T,Z)$, while $\mathcal{H}$ contains heating. At fixed density the net radiative gas cooling time is approximately $u/(\mathcal{C}-\mathcal{H})$ when the denominator is positive; a cooling flow closer to constant pressure must include enthalpy and compressional work. Comparing this time with the free-fall time of a uniform sphere determines whether pressure can be lost during gravitational contraction.

Hot ionized gas cools by bremsstrahlung and recombination; collisional excitation followed by photon escape supplies atomic and metal-line cooling at lower temperatures. molecular line cooling through rotational or vibrational lines and dust thermal emission become important in cold dense gas. Abundances, ionization state, density-dependent collisional de-excitation, and photon trapping all change the cooling function. Optically thick gas requires radiative transfer, not a simple local density-squared loss law.

Heating comes from photoionization, grain photoelectric heating driven by ultraviolet light, cosmic rays, shocks, turbulent dissipation and compression. photoionization heating deposits the photon energy above the ionization threshold as electron kinetic energy. A thermal equilibrium satisfies $\mathcal{H}=\mathcal{C}$; its stability depends on how the net loss changes under the relevant pressure or density constraint, not merely on equality. Different stable balances permit a multiphase interstellar medium. Spectral-line ratios constrain temperature and ionization, but interpreting them requires density diagnostics, interstellar extinction corrections and a radiation-field model.

The low-latitude regions of Earth absorb more incoming solar radiation than they emit as outgoing thermal radiation, whereas high latitudes have a net annual radiation deficit. The obliquity of solar incidence and the distribution of daylight create a strong latitude dependence of absorbed solar energy; emitted thermal radiation varies less strongly because terrestrial temperature contrasts are smaller. The terrestrial atmosphere and ocean therefore carry energy from the tropical surplus toward the polar deficits. Otherwise the low latitudes would keep warming and the high latitudes cooling until a different local radiative equilibrium was reached.

For a steady, zonally averaged energy balance, let $R(\varphi )$ be net downward radiation at the top of the terrestrial atmosphere and $H(\varphi )$ the total northward energy transport across latitude $\varphi$. A narrow spherical band and conservation of energy giveThe poleward direction is northward in the Northern Hemisphere and southward in the Southern Hemisphere. Seasonal internal energy storage modifies the instantaneous balance, but not the annual need for poleward redistribution.

Several circulation structures contribute in addition to the transient eddies produced by baroclinic instability of the middle latitudes:

The Brewer-Dobson circulation additionally redistributes energy in the stratosphere, with tropical ascent and extratropical descent, although most total atmospheric heat transport occurs lower down. An Eulerian mean middle-latitude Ferrel cell is thermally indirect and can oppose the poleward transport by eddies; listing every meridional circulation as a direct poleward heat carrier would be incorrect.

For $N$ distinguishable lattice sites occupied by $N_i$ mutually indistinguishable atoms of each species, the number of configurations isUsing Stirling's approximation,For one mole, $N=N_A$ and $k_B{N}_A=R$, soEquiatomic configurational entropy follows when the composition has $X_i=1/n$, givingThe reference is the unmixed configurational arrangement, and the model assumes random site occupation without short-range order. It counts configurational entropy, not all vibrational, electronic or magnetic entropy. A large number of elements does not by itself guarantee a stable single solid solution: competing thermodynamic phases and the enthalpy term also enter the Gibbs free energy.

For a spherical homogeneous nucleus, the capillarity approximation givesThe first term penalizes creating an interface; below the melting point the bulk free-energy change per volume $\Delta g_V$ is negative and favors growth. Near $T_m$, neglect the temperature dependence of the latent heat. Since the free-energy difference vanishes at $T_m$,Set the radius derivative to zero:The nonzero stationary point is a maximum, since $\Delta G^{\prime \prime }(r^{\ast })=-8\pi \gamma <0$. Subcritical nuclei lower free energy by shrinking; supercritical nuclei lower it by growing. The positive radius and its inverse-supercooling dependence follow from the negative freezing enthalpy. This derivation assumes an isotropic, size-independent interfacial energy and a bulk-like spherical nucleus; it is not a molecular-scale kinetic theory.

The energy-conserving self-similar blast wave requires a stage in which the original stellar radius, ejecta mass, ambient pressure and cooling have ceased to supply important scales. In particular, the swept-up mass must greatly exceed the ejecta mass; $R\gg R_{\ast }$ alone need not ensure this in a very dilute interstellar medium. For fixed explosion energy $E$, uniform ambient mass density ${\rho }_0$ and elapsed time $t$, dimensional analysis gives $R=C{E}^a{\rho }_0^b{t}^d$. Matching mass, length and time gives $a=1/5$, $b=-1/5$, $d=2/5$. Equivalently, a self-similar strong shock wave has $E\sim {\rho }_0{R}^3{\dot{R}}^2$, whose constant-energy solution has the same exponents:The dimensionless coefficient depends on $\gamma$ and the post-shock profiles; dimensional reasoning cannot fix it.

For completeness, obtain the Strong-shock Rankine-Hugoniot conditions in the shock frame. Let $v_1=V-u$ denote downstream speed relative to the outward-moving front. With negligible upstream pressure and enthalpy, conservation of mass, momentum and energy givesPut $q={\rho }_1/{\rho }_0$. Substituting $v_1=V/q$ into the last two equations and selecting the compressive, nontrivial branch yields $q=(\gamma +1)/(\gamma -1)$. HenceHere the symbol $u$, written as $u=2V/(\gamma +1)$ in what follows, is the laboratory-frame radial velocity; it is not the relative speed $v_1$.

Use the thin-shell approximation for a spherical blast wave. Set $W=4\pi R^3/3$ and $M_s={\rho }_{0}W$. Equating the swept-up mass to $4\pi R^{2}d{\rho }_1$ gives $d/R=(\gamma -1)/[3(\gamma +1)]$. The growing shell's momentum is $M_{s}u$. Since newly swept gas has zero laboratory momentum, the net force is the cavity pressure acting on the inner shell, to thin-shell accuracy:The pressure at the shock is already accounted for by the jump conditions and mass loading; subtracting $p_1$ as an additional external pressure would count the shock reaction twice. With $u=2V/(\gamma +1)$, the last equation becomesTherefore, for constant $\alpha$,

The shell volume is $M_s/{\rho }_1=W/q$. Its internal energy and kinetic energy areTreat the cavity volume as $W$ at the same thin-shell level, so its internal energy is $E_{c,\mathrm{int}}=\alpha p_{1}W/(\gamma -1)$. Their sum is proportional to $R^3{V}^2$ for constant $\alpha$ and $\gamma$. Constant total energy thus requires $3+6(\alpha -1)=0$, giving $\alpha =1/2$. ConsequentlyUsing $V=2R/(5t)$, this becomesFor $\gamma =5/3$, $d/R=1/12$ and the approximate energy fractions are $E_{s,\mathrm{kin}}/E=1/4$, $E_{s,\mathrm{int}}/E=1/4$, $E_{c,\mathrm{int}}/E=1/2$.

This is an approximate normalization, not an exact treatment of finite shell thickness. If instead one subtracts the shell volume and uses $W_c=W(1-1/q)$ for the cavity, while retaining the other shell assumptions, the energy coefficient becomes $2(2\gamma -1)/[(\gamma -1)(\gamma +1{)}^2]$ and ${\xi }_0^5=75(\gamma -1)(\gamma +1{)}^2/[32\pi (2\gamma -1)]$. The difference is an order-$d/R$ geometric correction of this crude shell model; a consistent exact coefficient requires the full self-similar profiles.

In the idealized mass budget all swept gas is assigned to the shell, so the cavity inertia is neglected. The hot cavity's residual mass density is not specified, so its kinetic energy cannot be assigned a unique numerical fraction from the stated data. If its low mass is $M_c$ and its velocity is of order $u$, then $E_{c,\mathrm{kin}}\sim M_c{u}^2/2$ andFor an illustrative uniform-density cavity with homologous velocity $v(r)=ur/R$, direct radial integration gives $E_{c,\mathrm{kin}}=3M_c{u}^2/10$, so the preceding fraction has an extra factor $3/5$. The cavity can carry substantial internal energy despite negligible kinetic energy because it is hot. Neglect of its inertia is justified in the swept-mass-dominated stage, not merely by its small initial radius.

The calculation describes the adiabatic expansion stage of a supernova remnant, giving an estimate of its age or explosion energy from its radius, expansion speed and ambient mass density. For example $t\simeq 2R/(5V)$, and $E\propto {\rho }_0{R}^3{V}^2$. Such blasts inject energy and momentum into the interstellar medium and redistribute stellar ejecta. The early ejecta-dominated stage, radiatively cooled late shell, nonuniform ambient medium, finite ambient sound speed, magnetic fields, nonspherical geometry and energy transfer to cosmic rays violate different assumptions. Even in the adiabatic spherical stage, the real profiles are not a uniform thin shell and a uniform-pressure cavity, so ${\xi }_0$ here is an estimate.

Write $q=bu$, $\rho ={\rho }_{\infty }$ and $r=\rho -{\rho }_s>0$. The first balance is depth-integrated momentum conservation. Its left side is the downstream increase of horizontal momentum flux. The wind contributes the applied stress $S$; the remaining term is the resultant hydrostatic force associated with the sloping base of the light layer. Because $\overline{\rho }-\rho <0$, that term opposes the wind when $b^{\prime }>0$.

For a strip of unit alongshore width and length $dx$, let $E=q^{\prime }$ be the entrained ocean-volume flux per unit area. Ocean entering the layer carries temperature $T_{\infty }$ and concentration $C_{\infty }$ but no horizontal momentum in this model. Take the reference liquid enthalpy at $T_{\infty }$ to be zero and setThe latent term is negative because ice has lower enthalpy than liquid at the same temperature. The incoming entrained water has zero relative enthalpy, and the outgoing minus incoming horizontal enthalpy flux is $[eq{]}^{\prime }dx$. Atmospheric heat loss therefore gives $[eq{]}^{\prime }=-f$. Similarly, solute conservation givesThese are the second and third balances. Their reference-flux form has already subtracted the heat and solute brought in by entrainment; those fluxes must not be subtracted again.

There is a dimensional normalization to make explicit. With the printed specific $c_p,L$, the physical heat-flux balance is $\rho [eq{]}^{\prime }=-F_{\mathrm{phys}}$. Thus $f=F_{\mathrm{phys}}/\rho$. The displayed exam equation uses its symbol $F$ for $f$, or implicitly works with density-normalized heat flux. The derivation below uses $f$ to keep these two meanings distinct.

With no incoming coastal jet or ice source, the conditions at $x=0$ areThe regular emerging layer also has $T\to T_{\infty }$, $\varphi \to 0$, $\overline{C}\to C_{\infty }$ and $b\to 0$. Prescribed nonzero inlet fluxes would replace these conditions and destroy the coast-origin power law. A zero-thickness coast does not justify independently prescribing an arbitrary nonzero velocity there.

Integration of the solute and enthalpy balances yields $\overline{C}=C_{\infty }$ wherever $q>0$, and $eq=-fx$. Local phase equilibrium now givesFor $\varphi \ll 1$, putTo leading order, $\varphi q=Kx$. The potential-energy closure implies $rg\varphi b{b}^{\prime }=S/4$, and substituting it in the momentum equation givesDividing the ice flux by this last relation gives $\varphi =(K/J)u$. If $b\propto x^p$, $u\propto x^q$, momentum gives $p+2q=1$, while the potential-energy balance gives $2p+q=1$. Thus $p=q=1/3$. Substitution fixes the coefficients:This is the dilute frazil polynya similarity solution. In particular, all three variables grow as $x^{1/3}$, $q$ grows as $x^{2/3}$, and the solution stops being dilute when $\varphi$ is no longer small. Density differences have been retained in buoyancy, but not in inertia or the normalized heat capacity.

To account for upward settling, one must distinguish retained suspension from crystals that leave it. A consistent depth-averaged extension has an upward exported ice-volume flux $R$ and an entrainment flux $E$. If all phases retain the same mean horizontal velocity within the layer, its balances becomeThe last term removes the momentum carried out by the ice; its density may be restored to ${\rho }_s$ outside the Boussinesq approximation. Consequentlyso one may no longer set $\overline{C}=C_{\infty }$. The local equilibrium relations are unchanged, but use the new bulk concentration. The frazil export enthalpy balance also shows why a simple replacement $-f\mapsto -f-LR$ has the wrong sign: exported ice carries negative relative enthalpy.

A slip speed $w$ requires a vertical particle-flux closure, for example $R=w\varphi$ for a well-mixed top boundary. A nonuniform suspension instead requires its surface concentration and buoyancy moment. The potential-energy closure must then include the gravitational energy released by rising light crystals, with scale $rgw\varphi b$, and any energy and buoyancy moment carried through the boundaries. For example, a well-mixed closure that transfers a fraction $\eta$ of this released power to entrainment and neglects additional boundary potential-energy transport would replace the original work balance byThe allocation $\eta$ and exported potential-energy flux are extra closure information, not supplied constants of the original model. The original work balance cannot be retained unchanged while treating settling as an additional loss of suspended ice. Phase-dependent horizontal speeds would also replace $q\varphi$ and $q(1-\varphi )$ by their separate phase fluxes.

It is important to compare total production, not merely the retained ice flux. LetIntegrated solute conservation gives $C(q-I)=C_{\infty }(q+R_{\ast })$, henceIntegrating the heat balance, and neglecting products of the small concentration departure with the exported flux, givesto dilute order, provided the total processed fraction $P/q$, not only the retained fraction $\varphi$, remains small. Thus upward settling reduces retained suspension, but does not increase the leading-order total production at fixed atmospheric heat loss. Beyond that order, efficient removal can reduce the subsequent sensible cooling of crystals and enhance mixing of ocean water; the usual expectation is a larger latent-heat fraction and hence greater production. Its amount, or an unconditional larger/smaller claim, needs the settling, mixing and surface-export closures just identified. At fixed $f$, energy conservation bounds the production by the available heat loss; it must not be inferred simply by observing fewer suspended crystals.

Mechanical balance and equipotentials. Let $\Phi$ be the Newtonian gravitational potential and $\varpi$ the cylindrical distance from the rotation axis. For rigid angular velocity $\Omega$, a fluid element at rest in the rotating frame has no Coriolis acceleration. The centrifugal acceleration is ${\Omega }^2\varpi \hat{\varpi }$, so hydrostatic equilibrium becomesUsing the Poisson equation for gravity and ${\nabla }^2(x^2+y^2)=4$ gives

On any regular connected equipotential surface, every tangent displacement satisfies $d\varphi =0$ and therefore $dP=0$. Locally $P=P(\varphi )$, with $dP/d\varphi =-\rho$. This makes $\rho$ a function of $\varphi$ as well. Equivalently, taking the curl of the hydrostatic equation gives $\nabla \rho \times \nabla \varphi =0$. Thus $P,\rho$ and ${\nabla }^2\varphi$ are constant on each surface. The last conclusion follows from the modified Poisson equation; it does not constrain the magnitude of the first derivative $|\nabla \varphi |$ to be constant there.

For uniform composition, an equation of state $P=P(\rho ,T)$ determines $T$ uniquely at fixed $P,\rho$ in the usual stellar regime. For the gas-plus-radiation equation, the derivative with respect to $T$ is positive. Hence $T=T(\varphi )$, and the opacity and radiative conductivity are also functions of $\varphi$.

Surface flux and the requested sketch. Define $f(\varphi )=-\chi dT/d\varphi$, positive for the outward radiative flux. ThenOn an outer equipotential surface, $f$ is constant and the outward normal is parallel to $\nabla \varphi$. Its effective gravity is $g=|\nabla \varphi |$, so $F_n=fg$. The Stefan–Boltzmann law $F_n={\sigma }_{\mathrm{SB}}{T}_e^4$ gives the radiative gravity-darkening law,A rapidly rotating star is oblate. Centrifugal support reduces the effective gravity near the equator, which is therefore cooler and dimmer; the poles are hotter. The original sketch below uses a subcritical Roche model of a uniformly rotating star to show this geometry and the corresponding temperature variation. The material temperature on an interior equipotential is not the same quantity as the emergent effective temperature defined by surface flux.

Why purely local radiative balance fails. Differentiating the flux givesOn one equipotential, $f,f^{\prime },{\nabla }^2\varphi$ and the proposed source $\rho ϵ(\rho ,T)$ are all constant, but $|\nabla \varphi {|}^2$ generally varies with latitude. Unless $f^{\prime }=0$ or there is another special cancellation, no single value of $\rho ϵ$ can match this flux divergence at all points on the surface. This is the radiative-equilibrium obstruction in a rotating barotropic star. Slow stellar meridional circulation can transport heat to compensate; it is distinct from abandoning hydrostatic balance at leading order.

Integrated energy balance with circulation. The steady continuity equation is $\nabla \cdot (\rho \mathbf{v})=0$. Combining the entropy–enthalpy relation with hydrostatic balance givesIntegrate the energy equation over $V$ and apply the divergence theorem:Uniform composition makes $h=h(\rho ,T)$ constant on the enclosing equipotential, and $\varphi$ is constant there by definition. Pull $(h+\varphi )$ out of the last integral; the remaining net mass flux is zero by steady continuity. Thus the equipotential luminosity conservation with stellar circulation isThis does not require $\mathbf{v}\cdot \hat{n}=0$ pointwise on $S$: inward and outward circulation may cancel in the net mass flux.

Finally, $f$ is constant on $S$ and the divergence theorem applied to the effective potential givesTherefore $-\chi dT/d\varphi =L/(4\pi Gm-2{\Omega }^{2}V)$. Since $dP/d\varphi =-\rho$,The mass $m$ is the mass enclosed by this surface, not necessarily the total stellar mass. The denominator is positive for the regular hydrostatic surfaces used here; near loss of centrifugal confinement the assumed stellar model must be reconsidered. The derivation uses slow circulation, retaining hydrostatic/equipotential thermodynamics to leading order while allowing a nonzero advective heat flux.

Use signed velocities in the shock frame; the Mach number depends on their squares. For a genuine nonradiative perfect gas shock wave, the Rankine-Hugoniot conditions for a perfect gas give a common mass flux $j={\rho }_L{u}_L={\rho }_R{u}_R$, a common momentum flux $p+\rho u^2$, and a common specific stagnation enthalpy $\gamma p/[(\gamma -1)\rho ]+u^2/2$. Here $e$ in the given equation of state is specific internal energy, despite the source's informal “energy density” wording.

Put $r={\rho }_R/{\rho }_L$, $P=p_R/p_L$ and $M^2={\rho }_L{u}_L^2/(\gamma p_L)$. Mass and momentum give $u_R=u_L/r$ and $P=1+\gamma M^2(1-1/r)$. Substituting these into the energy relation and clearing denominators givesThe $r=1$ factor is the no-jump solution. On the nontrivial shock wave branch,In particular $r-1=2(M^2-1)/[(\gamma -1)M^2+2]$. With $\gamma >1$, compression is therefore equivalent to $M_L^2>1$. The other Mach number isPositive pressure makes its denominator positive, and $M_R^2-1$ has the opposite sign to $M_L^2-1$. Thus a nontrivial physical shock wave has a supersonic side and a subsonic side; the entropy-admissible flow goes from the former to the latter.

The shock compression ratio also yields $u_L{u}_R=(\gamma -1)u_L^2/(\gamma +1)+2\gamma p_L/[(\gamma +1){\rho }_L]$. Rearranging produces the first velocity identity in the question. The pressure relation gives the second identity in the equivalent formThe numerator here contains mass density, as the original PDF shows; the converted TeX incorrectly replaces that mass density by pressure.

For reflection, the once-shocked gas is the common left state in both applications. Before rebound, its velocity in the shock frame is $a=u_s-U_{+}<0$ and the gas ahead has velocity $-U_{+}$; after rebound, it is $b=u_s+U_{-}>0$ and the gas next to the wall has velocity $U_{-}$. In both cases the right velocity minus the left velocity equals $-u_s$. Applying the first identity therefore gives the same quadratic equation,with roots $a$ and $b$. Their opposite signs distinguish the two roots, so $\boxed{(u_s-U_{+})(u_s+U_{-})=-\gamma p_s/{\rho }_s}$.

Write $q_0=p_0/p_s$, $q_1=p_1/p_s$. The pressure identity applied to each shock wave givesSquaring the root-product relation and eliminating the velocities yieldsand hence the normal shock reflection at a rigid wall relationTaking $p_0/p_s\to 0$ gives $\boxed{p_1/p_s\to (3\gamma -1)/(\gamma -1)}$; for a monatomic gas this limiting amplification is six.