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An adiabatic fluid perturbation preserves the specific entropy of each displaced parcel. For a perfect gas, mass conservation gives $\Delta \rho =-\rho \nabla \cdot \mathbf{\xi }$ and the adiabatic equation of state gives $\Delta p=\gamma p\Delta \rho /\rho$. The corresponding Eulerian perturbation of a fluid variable is $\delta p=-\gamma p\nabla \cdot \mathbf{\xi }-\mathbf{\xi }\cdot \nabla p$. An isothermal equilibrium may still have adiabatic perturbations, because the thermodynamic constraint on displaced parcels differs from the equilibrium temperature profile.

For a steady axisymmetric compressible fluid flow, mass conservation permits this representation. The mass flow through an annulus between two stream surfaces is $2\pi \Delta \psi$. In a steady adiabatic wind without magnetic forces, the specific entropy, specific angular momentum and Bernoulli function are locally functions of $\psi$ on connected regular stream surfaces.

For an inviscid barotropic fluid with only potential body forces, the vorticity $\mathbf{\omega }=\nabla \times \mathbf{u}$ obeysThe pressure force is a gradient because ${\rho }^{-1}\nabla p=\nabla {\int }^{\rho }{p}^{\prime }(s)ds/s$. A spatially uniform specific entropy makes a fixed-composition fluid barotropic. Merely having $Ds/Dt=0$ does not exclude baroclinic vorticity generation from spatial entropy gradients.

A contact discontinuity in one-dimensional gas dynamics moves with the common fluid velocity $s=u_L=u_R$ and has continuous pressure, while mass density and specific entropy can jump. The mass jump condition $s[\rho ]=[\rho u]$ is automatic with common $u=s$; momentum conservation $s[\rho u]=[\rho u^2+p]$ then requires $[p]=0$. The energy jump condition is also satisfied. Thus a contact is a material interface, not a compression shock. A smooth self-similar positive-density flow cannot spread this zero-speed-relative-to-the-fluid family into a fan: $u=x/t$ on an interval would require both $du/d(x/t)=1$ and, from continuity, $du/d(x/t)=0$.

For a monatomic fully convective ideal gas star, uniform specific entropy gives $P=K{T}^{5/2}$ with $K\propto {\mu }^{-5/2}{M}^{-1/2}{R}^{-3/2}$. Combining this with the stellar surface boundary condition $\kappa P=2g/3$ and $\kappa ={\kappa }_0{\rho }^a{T}^b$ givesAll composition comparisons here hold ${\kappa }_0$ fixed. For $a=1,b=4$, $T_e^8\propto {\mu }^4{M}^{2}R$, and the Stefan–Boltzmann law gives $L\propto {\mu }^{2}M{R}^{5/2}$. The stellar virial theorem and Kelvin-Helmholtz contraction then give $R^{-7/2}-R_0^{-7/2}\propto {\mu }^2{M}^{-1}(t-t_0)$, with a positive coefficient. Once the initial-radius term is negligible, $R\propto {\mu }^{-4/7}{M}^{2/7}(t-t_0{)}^{-2/7}$.

A uniform-density perfect-gas sphere supported by decreasing pressure has buoyancy frequency squared $N^2=-2{\omega }_d^2{r}^2/[\gamma (R^2-r^2)]$ in the interior. Its specific entropy decreases outward and adiabatic displacements are buoyantly unstable. For the nonradial polynomial family, the product of the two normalized squared mode frequencies is $-l(l+1)$, displaying one growing branch even when radial compression modes are stable.

An adverse radial specific entropy gradient can drive an oscillatory instability through the finite lag supplied by thermal conduction or relaxation. This differs from a viscous-convective instability, in which viscosity changes rotational stabilization.

A stable stratification gives a restoring buoyancy force and positive squared buoyancy frequency for an adiabatically displaced parcel. An adverse specific entropy gradient can instead drive convection; other forces such as rotation can modify the full stability criterion.

Let $h_c=\mathbf{u}\cdot \mathbf{B}$ be the cross-helicity density. The ideal magnetohydrodynamic induction equation implies $D\mathbf{B}/Dt=(\mathbf{B}\cdot \nabla )\mathbf{u}-\mathbf{B}\nabla \cdot \mathbf{u}$. Dotting the momentum equation with $\mathbf{B}$ eliminates the Lorentz force density, since it is perpendicular to the magnetic field. The material derivative of $h_c$ is thereforeUsing $dh=Tds+dp/\rho$ for specific enthalpy $h$ and specific entropy $s$, together with Gauss's law for magnetism, converts this toThe source vanishes when magnetic field lines lie in constant-specific entropy surfaces, in particular in a homentropic flow. Integrating over a volume conserves its total cross-helicity only if the boundary flux also vanishes. This local law is derived in Ogilvie's astrophysical fluid dynamics notes.

With negligible nuclear or external heating, the stellar structure equations give $L_{\mathrm{int}}=-{\int }_0^{M}T\partial s/\partial tdm$. A nearly isentropic convective interior therefore has $\dot{s}=-L_{\mathrm{int}}/\int Tdm<0$ whenever it radiates. Decreasing specific entropy does not require temperature at every depth to decrease immediately: contraction of an ideal gas can raise its central temperature while total energy is lost.

For the pressure-density Hugoniot relation for a perfect gas, the change in specific entropy is $[s]/c_v=\log P-\gamma \log D(P)$, with $c_v$ the specific heat capacity at constant volume. Its derivative isIt is positive for a compressive normal shock wave with $P>1$. For a weak shock with $P=1+\delta$, integrating the leading term givesThus the entropy production is cubic in the small pressure jump, even though the mass density and temperature changes already appear at first order.

An ice bath can absorb heat while its temperature barely changes: the heat converts ice into water. The latent heat of fusion quantifies this melting budget. For a pure substance at its equilibrium melting temperature $T_m(p)$ and specified pressure $p$, define it here per unit mass:Here $h_l$ and $h_s$ are the specific enthalpies of the coexisting bulk liquid and solid, and $H$ is enthalpy. For ordinary melting, $L>0$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$; an amount $m$ has a total fusion enthalpy $mL$ in joules. A molar fusion enthalpy instead has units $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ and equals $ML$, where $M=m/n$ is mass divided by amount of substance in $\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. Always check the normalization before inserting a tabulated value into an energy balance.^{[ref] [ref]}

For a closed system converting completely from solid to liquid at fixed pressure, with only pressure-volume work and negligible changes in bulk kinetic energy and potential energy, the first law of thermodynamics gives $Q_p=\Delta H=mL$. Both endpoints are at $T_m(p)$: warming initially colder ice or subsequently warming the resulting water requires additional sensible heat. Since $H=U+pV$, the corresponding specific internal energy jump is $\Delta u=L-p\Delta v$, where $v=V/m$; $L$ includes the work associated with the change in volume. Heat is transferred energy, whereas enthalpy is a state function. ^[ref]

Let $x=m_l/m$ be the liquid phase fraction by mass, rather than by volume. Neglecting interfacial contributions to enthalpy, equilibrium phase coexistence at fixed pressure givesIn the model above, reaching $T_m$ takes $2.10\mathrm{kJ}$ of sensible heat; the plateau then absorbs $33.40\mathrm{kJ}$, ending at a cumulative $35.50\mathrm{kJ}$. At $18.80\mathrm{kJ}$, half the mass is liquid and the temperature is still $T_m$. The final warming takes another $4.18\mathrm{kJ}$. The rounded $L$ and $c_l$ are inspired by the introductory ice-melting calorimetry values in Case Western's demonstration; constant properties and exact equilibrium throughout are stipulated model assumptions. ^[ref]

Within one thermodynamic phase, specific heat capacity at constant pressure is $c_p=(\partial h/\partial T{)}_p$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$; a sample's heat capacity at constant pressure is $C_p=m{c}_p$, in $\mathrm{J}{\mathrm{K}}^{-1}$. The plateau is a finite enthalpy change at zero temperature change through varying $x$, so neither a finite single-phase $c_p$ nor $Q=m{c}_p\Delta T$ represents it. The heating curve needs separate sensible heat intervals and a fusion contribution. ^[ref]

The plateau assumes a pure substance, a stable bulk solid-liquid phase coexistence pair, fixed pressure, and sufficiently slow, uniform heating to remain near thermodynamic equilibrium. The total mass and composition must remain fixed, with negligible evaporation and no other phase transition. A fast heater can produce temperature gradients, so a local thermometer need not read the equilibrium $T_m$. Neither $T_m$ nor $L$ is universal across pressures or compositions: common-pressure melting-point slope describes the pressure dependence of $T_m$, while mixtures can melt over a temperature range as their coexisting compositions change. Their liquidus must be consulted instead of assuming a pure-substance plateau. ^[ref]

This makes ice useful as a temperature buffer: under these conditions it can absorb $m_{s}L$ before the remaining solid mass $m_s$ disappears, after which the water warms. If a stipulated constant net input power $P$ goes entirely into melting, that buffer lasts $m_{s}L/P$. Reversing the equilibrium conversion releases the same fusion enthalpy during freezing. The beautiful thermodynamic connection is $L=T_m(s_l-s_s)$, with $s$ the specific entropy and $T_m$ absolute temperature: equal Gibbs free energies at phase coexistence allow different enthalpies and entropies. The nearly stationary thermometer hides a finite, measurable change of state. ^[ref]

In a radiative shock, a narrow dissipative jump first converts upstream kinetic energy into internal energy. The immediate strong adiabatic compression is $(\gamma +1)/(\gamma -1)$, and the immediate temperature is the result of part 1(c). Downstream gas then radiates, loses specific enthalpy and compresses further. Its shock cooling length is of order $u_2{t}_{\mathrm{cool}}$, with the immediate downstream speed used only as an estimate because the gas slows as it cools.

For the idealization of equal far-upstream and cooled downstream temperatures, unchanged composition and negligible magnetic stress, write $p=c_T^2\rho$. Conserved mass and normal momentum flux across the whole layer giveWith $r={\rho }_f/{\rho }_1=u_1/u_f$, the nontrivial isothermal shock root isIt can greatly exceed the adiabatic strong-shock compression. An upstream adiabatic sound speed instead defines ${\mathcal{M}}_{\mathrm{ad},1}=u_1/(\sqrt{\gamma }{c}_T)$, so the same result is $r=\gamma {\mathcal{M}}_{\mathrm{ad},1}^2$; the two Mach numbers must not be confused. The radiated energy flux is ${\rho }_1{u}_1(u_1^2-u_f^2)/2$ when the two endpoint specific enthalpies agree. The gas's specific entropy rises at the initial jump and falls during cooling, while the escaping radiation carries entropy away. A transverse frozen-in magnetic field, or an imposed temperature floor, can limit the compression. Strongly cooling shocks can therefore collect thin dense shells without violating total conservation of energy.

In a spherical self-gravitating equilibrium with positive mass density, $d\psi /dr=GM(r)/r^2>0$ away from the center. Thus $r$ can locally be replaced by $\psi$, so both $p(r)$ and $\rho (r)$ can be expressed as functions of $\psi$. Hydrostatic equilibrium further givesSince $p$ decreases outward wherever $\rho >0$, eliminating $\psi$ gives a barotropic equation of state along this equilibrium. This is a relation describing its structure; it does not require all thermodynamic perturbations of an arbitrary material to be barotropic.

For $p=K{\rho }^{1+1/n}$, integrate the polytropic enthalpy balance $dp/\rho =-d\psi$:For a finite-radius zero-pressure boundary with $K>0$ and $n>0$, $\rho (R)=0$ and therefore ${\psi }_T=\psi (R)$. With the gravitational potential zero at infinity, an isolated sphere has ${\psi }_T=-GM/R$. It is an integration constant tied to that boundary, not necessarily a zero of the gravitational potential.

Put $\theta =(\rho /{\rho }_c{)}^{1/n}$ and $\xi =\alpha r$, with ${\alpha }^2=4\pi G{\rho }_c^{1-1/n}/[(n+1)K]$. The preceding relation gives $\psi ={\psi }_T-(n+1)K{\rho }_c^{1/n}\theta$. Substitution into the spherical gravitational Poisson equation yields the Lane-Emden equationLet ${\xi }_1$ be the first zero of the regular Lane-Emden function and $Y_n=-{\xi }_1^2{\theta }^{\prime }({\xi }_1)>0$. Then the Lane-Emden mass formula and radius areFor a common fixed polytropic index with a finite first zero, their scalings areThe usual regular zero-pressure models have a finite first zero for $0<n<5$; the $n=5$ model has infinite radius, so a finite-radius formula cannot simply be applied to it.

For an ideal gas of fixed composition, $T_c\propto K{\rho }_c^{1/n}$. Consequently homologous polytropes at fixed central temperature have $K\propto {\rho }_c^{-1/n}$ and both $M,R\propto {\rho }_c^{-1/2}$, recovering $M\propto R$. The common central temperature, composition and dimensionless structure are the relevant assumptions. Keeping $K$ fixed instead gives the different polytropic mass-radius relation, and no finite index makes its generic fixed-$K$ exponent exactly one.

For adiabatic monatomic hydrogen, $\gamma =5/3$ and $n=3/2$. At a common fixed specific entropy, hence common $K$, the same relations giveHere $\bar{\rho }=3M/(4\pi R^3)$, and the omitted constant in the last relation depends on $K$. A population with different cloud entropies need not follow that mass sequence.

Efficient radiative cooling in molecular clouds prevents compression from forcing the temperature upward as steeply as in an adiabatic monatomic gas. Rotational transitions of molecules and radiation by dust provide low-temperature cooling channels; which dominates depends on composition, opacity and temperature. It is therefore possible to assemble dense massive cold clouds while disposing of compressional heat. This explains the direction of the idealized comparison, rather than proving a universal cloud-mass cutoff.

For a common isothermal sound speed $c_T$, a spherical isothermal equilibrium has $GM/R$ equal to $c_T^2$ times a dimensionless structural factor. Holding that factor fixed gives the homologous isothermal mass-radius relationFor example, the singular isothermal sphere has $M(R)=2c_T^{2}R/G$. A regular finite cloud requires a pressure-confined isothermal sphere or another outer boundary: an unconfined isothermal sphere does not end at a finite vacuum radius. Equal temperature alone, while allowing arbitrary truncation or density contrast, does not make the coefficient universal.

The Mach number is $\mathcal{M}=u/c_s$, comparing the flow speed relative to a disturbance with the local sound speed. For a body moving supersonically, spherical acoustic disturbances emitted earlier have a tangent envelope with Mach cone half-angle $\sin \theta =1/\mathcal{M}$. Signals cannot propagate arbitrarily far upstream. Compressive characteristics can converge and steepen into a shock wave, within which microscopic dissipation prevents a multivalued fluid state; supersonic motion alone does not require every flow to contain a shock.

Work in the shock frame with steady normal velocities $u_1,u_2>0$. Integrating conservation of mass, normal momentum and total energy over a thin pillbox, neglecting gravity across the thin layer, gives the Rankine-Hugoniot conditions for a perfect gas:The energy equation includes pressure work through specific enthalpy. The shock is adiabatic if negligible heat escapes the layer, but its irreversible dissipation means it is not isentropic.

Put $\chi ={\rho }_2/{\rho }_1=u_1/u_2$ and ${\mathcal{M}}_1=u_1/\sqrt{\gamma p_1/{\rho }_1}$. Momentum conservation gives $p_2=p_1+{\rho }_1{u}_1^2(1-1/\chi )$. Substitution in the energy equation produces the unshocked root $\chi =1$ and the compressive rootSince ${\rho }_1{u}_1^2/p_1=\gamma {\mathcal{M}}_1^2$, the strong-shock limit has negligible upstream thermal pressure compared with ram pressure. The Strong-shock Rankine-Hugoniot conditions giveFor unchanged composition, the ideal gas law implies $T_2/T_1=(p_2/p_1)/\chi$, soFor $\gamma =5/3$, the compression tends to four and $T_2\simeq 3\bar{m}{u}_1^2/(16k_B)$, where $\bar{m}$ is mean particle mass.

With constant specific heat $c_v$, entropy production in a perfect-gas shock is measured by the change in specific entropyThe strong-shock expression is thereforeThe exact jump is positive for every compressive shock with ${\mathcal{M}}_1>1$, in accordance with the Second law of thermodynamics; the asymptotic expression is only for ${\mathcal{M}}_1\gg 1$. There is a normalization issue in the printed description: this logarithmic formula with constant $c_v$ describes entropy per unit mass. The true entropy density is $\rho s$, so its difference is ${\rho }_2{s}_2-{\rho }_1{s}_1$, rather than the same logarithmic jump with a constant heat capacity per unit volume. A material parcel's irreversible entropy increase should be assessed using $\Delta s$.

Downstream gas is denser and hotter. For optically thin collisional emission the loss per unit volume is typically $n_e{n}_i\Lambda (T)$, whereas stored thermal energy scales approximately as $nT$. Compression can therefore reduce the radiative gas cooling time; excitation and ionization at the higher temperature can open additional cooling channels. The actual efficiency still depends on the astrophysical cooling function, composition and optical depth. In a radiative shock with separated scales, the temperature first rises across a narrow dissipative layer, then decreases through a much broader cooling region of characteristic length ${\ell }_{\mathrm{cool}}\sim u_2{t}_{\mathrm{cool}}$. Subsonic cooling usually brings further compression with approximately constant pressure. The immediate microscopic shock thickness and the macroscopic cooling length are distinct.

Let state $f$ be the fully cooled gas, with $T_f=T_1$ and the same composition. Set $a^2=p_1/{\rho }_1$, the isothermal sound speed squared. Across the entire shock plus cooling layer, mass and momentum are still conserved, but gas energy alone is not. With ${\chi }_f={\rho }_f/{\rho }_1$,The nontrivial isothermal shock therefore hasIf the upstream Mach number were defined with $a$ instead, this would be ${\rho }_f/{\rho }_1={\mathcal{M}}_{\mathrm{iso}}^2$. Retaining the distinction avoids losing the factor $\gamma$.

The gas's specific entropy rises irreversibly at the hot shock and then decreases as escaping radiation removes heat. Since the final temperature equals the initial temperature,This decrease for the gas is consistent with the Second law of thermodynamics: the emitted radiation and its eventual surroundings carry entropy away. It does not make the microscopic compressive shock reversible. The figure illustrates both the hot entropy jump and the subsequent cooling; its spatial cooling profile is schematic because no cooling law has been specified.

Let $z$ increase upwards and let $g>0$ denote the downward gravitational acceleration. The environment obeys hydrostatic equilibrium, $dp/dz=-\rho g$. Displace a small parcel by $\zeta$ while it remains in pressure balance with its surroundings and exchanges negligible heat. For fixed composition and adiabatic index $\gamma$, its density change at constant specific entropy satisfies $d{\rho }_{\mathrm{par}}/dp=\rho /(\gamma p)$. At the displaced position, its density excess over the environment is thereforeThe buoyancy acceleration is $\ddot{\zeta }\simeq -g\Delta \rho /\rho$. HenceStability requires $N^2>0$, equivalentlyAn upward-displaced parcel is then denser than its surroundings and falls back; a downward-displaced parcel is lighter and rises. Equality gives neutral stability in this parcel model. For a perfect gas with constant composition, $ds=c_{v}d\ln (p/{\rho }^{\gamma })$ gives $N^2=(g/c_p)ds/dz$, so a rising specific entropy is stable. This is the parcel form of the Schwarzschild criterion. Rapid pressure equilibration, negligible heat diffusion during displacement and no composition-driven buoyancy are the assumptions behind this test.

Work in the rest frame of a thin planar shock wave, with normal upstream and downstream velocities $u_1,u_2>0$. Assume a steady flow, negligible gravitational change across its thickness, no radiative energy loss, and negligible viscous stress and heat flux at the two remote faces of a control volume. Dissipation within the shock is permitted and is essential to the physical discontinuity.

Integrating mass conservation through that volume gives equal mass flux per unit area:The dimensions of $j$ are $\mathrm{kg}{\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$. The normal momentum entering minus that leaving must balance the pressure forces. Equivalently, the conserved normal momentum flux is $p+\rho u^2$, givingFor conservation of energy, matter advects kinetic and internal energy while pressure performs work at rate $pu$ per unit area. Writing $ϵ$ for specific internal energy, the energy flux is $\rho u(u^2/2+ϵ)+pu$. With specific enthalpy $h=ϵ+p/\rho$, its conservation givesThe division by $j$ is valid for a genuine through-flow shock. Together these are the three Rankine-Hugoniot conditions. The conserved energy flux has units $\mathrm{W}{\mathrm{m}}^{-2}$, and both momentum fluxes have units of pressure.

For a calorically perfect ideal gas, $ϵ=p/[(\gamma -1)\rho ]$ and $h=\gamma p/[(\gamma -1)\rho ]$. The three jump conditions can be combined explicitly. Put $r={\rho }_2/{\rho }_1$, so $u_2=u_1/r$ andSubstitute these into the enthalpy equation and use ${\mathcal{M}}_1^2={\rho }_1{u}_1^2/(\gamma p_1)$, with the upstream adiabatic sound speed. Besides the trivial no-shock root $r=1$, the physical compression root isFor unchanged composition, $T_2/T_1=(p_2/p_1)/r$. A compressive shock has ${\mathcal{M}}_1>1$ and increases the specific entropy. Thus an adiabatic shock conserves total energy but is not an isentropic process.

The local conservation laws. The first equation is the radial Euler equation: the material acceleration $Du/Dt={\partial }_{t}u+u{\partial }_{r}u$ is caused by the pressure gradient force per unit mass, $-{\partial }_{r}p/\rho$. It neglects gravity and viscosity. The second equation is mass conservation in spherical coordinates,The term $2\rho u/r$ appears because the area of a spherical surface grows as $r^2$.

To obtain the third equation, let $e=p/[(\gamma -1)\rho ]$ be the specific internal energy of an ideal gas with constant heat capacity ratio $\gamma$. The first law of thermodynamics for a smooth adiabatically moving element isDifferentiating $e$ and collecting terms givesand henceThus the entropy parameter $p/{\rho }^{\gamma }$ is conserved along each smooth pathline; for fixed composition, specific entropy differs from $c_v\ln (p/{\rho }^{\gamma })$ only by a constant. This is local isentropic flow behind the shock. A physical shock wave itself generates entropy and must be treated with jump conditions, rather than by applying this smooth equation across the discontinuity.

The front-radius scaling. In the self-similar blast wave approximation there is no scale set by ambient pressure, cooling, gravity, or the initial source radius. Write $R\propto E^a{\rho }_0^b{t}^d$. Using $[E]=\mathsf{M}{\mathsf{L}}^2{\mathsf{T}}^{-2}$ and $[{\rho }_0]=\mathsf{M}{\mathsf{L}}^{-3}$, dimensional analysis givesThus $a=1/5$, $b=-1/5$, $d=2/5$ andThe dimensionless coefficient ${\xi }_0$ depends on $\gamma$ and the resolved similarity profiles; dimensions alone do not fix it. Define the length scale $\ell (t)=(E{t}^2/{\rho }_0{)}^{1/5}$, so $\xi =r/\ell$ and the front is at the fixed coordinate $\xi ={\xi }_0$.

Kinetic energy. Substitute $\rho ={\rho }_0{f}_{\rho }(\xi )$ and $u=(r/t)f_u(\xi )$ into the volume integral. Since $r=\ell \xi$ and $dr=\ell d\xi$,But ${\rho }_0{\ell }^5/t^2=E$, so the first of the spherical blast-wave similarity energy integrals isThe conclusion requires a fixed, integrable similarity profile, as in the finite-energy blast solution. Expansion increases the swept-up mass while reducing the characteristic squared speed, leaving this product constant.

Pressure and internal energy. The derivatives of the similarity variable areWriting a prime for differentiation with respect to $\xi$, the material acceleration becomesThe radial momentum equation therefore requires a pressure gradient on the scale ${\rho }_0\ell /t^2$. Integrating over radius gives pressure similarity in a spherical blast wave:whereThe prefactor has units $\mathrm{kg}{\mathrm{m}}^{-1}{\mathrm{s}}^{-2}$, as required for pressure. Strictly, the momentum equation alone permits an additional pressure uniform in space and arbitrary in time. In the scale-free strong-blast solution, shock matching and the remaining thermodynamic equation fix that freedom with the same similarity scaling; any allowed constant of integration is included in $f_p$. An independently imposed external pressure would introduce another scale and invalidate the simple dimensional ansatz.

For constant $\gamma >1$, the internal energy is consequentlyIf $E$ is the total excess explosion energy, its normalization isThe two constant fractions need not be equal. No explicit solution for $f_p$ is needed to establish their time independence.

Efficient convection keeps the interior at nearly uniform specific entropy. For the fully ionized monatomic gas, the heat capacity ratio is $5/3$, so the adiabatic process gives $P=K_{\rho }{\rho }^{5/3}$, with $K_{\rho }$ spatially constant. The ideal gas relation $P=\mathcal{R}\rho T/\mu$ then impliesThe constant is spatial, not temporal: loss of specific entropy during contraction changes it. This is a stellar polytrope with polytropic index $n=3/2$. We neglect radiation pressure and take the thin photosphere as an atmospheric matching layer rather than a significant fraction of the stellar mass.

For a fixed dimensionless profile of the Lane-Emden equation, the stellar homology relations have the formwhere $C_{\rho },C_P$ are positive dimensionless structure constants. More explicitly, write $\rho (r)={\rho }_{c}f(x)$ and $m(r)=Mh(x)$ with $x=r/R$ and fixed dimensionless profiles. Mass conservation gives $M=4\pi {\rho }_c{R}^3{\int }_0^{1}f(x)x^{2}dx$. Integrating the hydrostatic pressure support equation, with negligible surface pressure on the interior scale, gives $P_c=(GM{\rho }_c/R){\int }_0^{1}h(x)f(x)x^{-2}dx$. Both integrals are fixed positive numbers, proving the two central scalings. Substituting these relations into $K=P_c/T_c^{5/2}$ givesThus the same polytropic index fixes $K_0$ throughout this homologous sequence, while the dimensional stellar polytrope constant changes.

At the photosphere, whose optical depth is $2/3$, identify the temperature with the effective temperature $T_e$. The ideal gas relation and the interior adiabat give ${\rho }_{\mathrm{ph}}=\mu K{T}_e^{3/2}/\mathcal{R}$. With the prescribed opacity and stellar surface boundary condition,Eliminating $K$ therefore provesUsing the Stefan–Boltzmann law, with $\sigma =ac/4$, gives the luminosity

The gravitational energy of a stellar polytrope is $\Omega =-6G{M}^2/(7R)$ for $n=3/2$. The stellar virial theorem for a monatomic gas gives $2U+\Omega =0$, and hence the total energy is $E=U+\Omega =-3G{M}^2/(7R)$. With no appreciable stellar nuclear fusion or accretion, energy conservation givesCombining the two expressions for $L$ yieldsMultiplication by $-7R^{-9/2}/2$ and integration proves the contraction law with density-linear fourth-power stellar opacity:When the accumulated contraction term dominates the initial-radius term, $R\propto {\mu }^{-4/7}{M}^{2/7}(t-t_0{)}^{-2/7}$. Substitution into the central ideal gas relation giveswhich has the stated $t^{2/7}$ behavior when the time origin is negligible. This late-time approximation still requires the assumed fully convective star, fixed composition and opacity law to remain applicable.

Let $b=\frac{1}{2}{u}^2+\Phi +w$. Because the specific entropy is constant along the horizontal flow, the specific enthalpy identity gives $\mathbf{u}\cdot \nabla w={\rho }^{-1}\mathbf{u}\cdot \nabla p$. Dotting the ideal magnetohydrodynamic momentum equation with $\mathbf{u}$ therefore givesOn the field-aligned horizontal branch, $\mathbf{u}\times \mathbf{B}=U{\mathbf{e}}_z\times \mathbf{B}$. Directly evaluating the horizontal components of the curl givesThe scalar triple-product identity consequently makes the magnetic work $U{\mathbf{B}}_h\cdot \nabla B_z/{\mu }_0$. Meanwhile $\rho \mathbf{u}\cdot \nabla b=k{\mathbf{B}}_h\cdot \nabla b$. The previously established field-line constancy of $U,k$ now gives the planar ideal-MHD Bernoulli invariantThis is conservation of total energy transported along a magnetic field line, including electromagnetic work. To see its flux interpretation, the ideal magnetohydrodynamic energy conservation law has fluxThe second term is the Poynting vector, since $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$. Substitution of $\mathbf{u}=k\mathbf{B}/\rho +U{\mathbf{e}}_z$ givesThe steady divergence of this horizontal energy flux is zero. Magnetic stress can change the ordinary hydrodynamic quantity $b$ in the Bernoulli equation by doing work; the term $-U{B}_z/({\mu }_{0}k)$ accounts for the accompanying Poynting flux. When $U=0$, the velocity is completely field-aligned, $\mathbf{E}=0$, and the usual hydrodynamic Bernoulli quantity is constant. These conclusions require the field-aligned branch and $k\ne 0$; the arbitrary constant-$E_z$ flow from the first part is not covered by this particular invariant.