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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.

The entropy form of dry potential temperature follows from its parcel definition. The potential temperature is the temperature a parcel would reach if brought reversibly and adiabatically to a reference pressure $p_0$. For dry air of fixed composition, treated as an ideal gas with constant specific heat capacity,Per unit mass, the first law of thermodynamics, together with $\alpha =R_{d}T/p$ for specific volume, gives $\delta q=c_{p}dT-\alpha dp$. The reversible form of the Second law of thermodynamics then givesHence dry reversible adiabatic motion has $D\theta /Dt=0$. Pressure changes cause expansion cooling or compression warming, while the pressure factor in theta compensates exactly. Adiabaticity alone is insufficient if irreversible mixing or viscous dissipation produces entropy. Condensation or freezing supplies latent heating, so ordinary dry potential temperature need not be conserved in a moist adiabatically moving parcel.

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.

For nonrelativistic particles the energy density is dominated by rest energy, while their isotropic kinetic pressure is approximately $p_m=nm⟨v^2⟩/3$. ThereforeThe matter pressure is negligible in this approximation, including in the gravitational combination ${\rho }_m+3p_m/c^2$. For isotropic radiation in cosmology, $E=pc$ implies one third of the energy density in each diagonal spatial stress:

Because the two fluids do not exchange energy or momentum, they are separately conserved cosmological fluids: ${\nabla }_{\alpha }{T}_i^{\alpha \beta }=0$ for each, rather than merely conservation of their sum. In a comoving volume proportional to $a^3$, the first law of thermodynamics givesEquivalently this follows by projecting the individual stress-energy conservation equation along the common comoving four-velocity. Differentiating and dividing by ${\rho }_i{c}^2{a}^3$ yieldsFor constant equation-of-state parameter $w_i$,The extra power for radiation in cosmology represents the cosmological redshift of photon energies in addition to number dilution. Differentiating the total Friedmann equation gives only the total continuity equation; the separate equations also use the stated absence of interactions between components.

For a perfect fluid in general relativity, let $\rho$ be rest-frame total energy density divided by $c^2$, $P$ its isotropic pressure, and $U^{\mu }$ its four-velocity, normalized by $U^{\mu }{U}_{\mu }=-c^2$. ThenThe contracted Bianchi identity and the Einstein field equations give stress-energy conservation, ${\nabla }_{\mu }{T}^{\mu \nu }=0$, for a constant cosmological constant. Contract with $U_{\nu }$ and use the normalization to obtainIn comoving coordinates $U^{\mu }=(1,0,0,0)$ and the covariant divergence is ${\nabla }_{\mu }{U}^{\mu }=3\dot{R}/R=3H$, where a dot is a cosmic time derivative and $H$ the Hubble parameter. ThereforeFor a fixed region in comoving coordinates, let $V(t)=V_{\ast }{R}^3(t)$ be its physical volume, where $V_{\ast }$ is a fixed coordinate-volume integral, and let $E=\rho c^{2}V$. Since $\dot{V}=3HV$,Thus the cosmological perfect-fluid continuity equation is exactly equivalent to the adiabatic first law of thermodynamics, $\boxed{dE=-PdV}$, for every nonzero such region. Conversely, expanding that first law of thermodynamics and dividing by $c^{2}V$ recovers the same equation. No heat or particle flux crosses the comoving boundary; interacting components separately need source terms, although the total still obeys this conservation equation.

There is a sign inconsistency in the supplied formula sheet worth isolating. Its Riemann curvature tensor and Ricci tensor conventions, together with its stated $-+++$ metric signature, give $G_{tt}=-3H^2$ for $R(t)=e^{Ht}$ with constant $H$. Its vacuum Einstein field equations with $+\Lambda g_{\mu \nu }$ then require $H^2=-\Lambda c^2/3$, whereas its printed vacuum Friedmann equation gives $H^2=+\Lambda c^2/3$. For example, $H>0$, $\rho =P=0$ and $\Lambda =3H^2/c^2$ satisfy the latter but leave the former's $tt$ component equal to $-6H^2$. Changing the geometric cosmological constant term to $-\Lambda g_{\mu \nu }$ repairs this mismatch. The conservation derivation is unaffected; below take the geometric cosmological constant to be zero and include any vacuum energy as a fluid, so no ambiguous sign enters.

There are two first-order effects: an intrinsic temperature change from adiabatic compression, and the gravitational redshift of the emerging photons. Evaluate the excess Newtonian gravitational potential at the boundary of the spherical fluctuation, with the exterior reference potential subtracted:In the weak-field approximation, gravitational time dilation gives $d{\tau }_{\mathrm{local}}/dt\simeq 1+\delta \Phi /c^2$. A local emitter's successive wave crests therefore arrive farther apart at a reference observer outside the potential well. Their frequency changes by $\delta \nu /\nu =\delta \Phi /c^2<0$. Since a redshifted black-body radiation spectrum has its temperature scaled by the same factor as its frequency, the gravitational contribution is

For the intrinsic effect, compare the local scale factor $R+\Delta R$ with the unperturbed value $R$ at the reference epoch. An equilibrium photon gas has energy density $u=a_{\mathrm{rad}}{T}^4$ and pressure $p=u/3$. For adiabatic compression, the first law of thermodynamics applied to a comoving volume $V\propto R^3$ givesThus $TR$ is constant along this adiabatic process, and the initial intrinsic contribution is $\Delta T/T=-\Delta R/R$. The smaller local expansion associated with the fluctuation leaves its radiation hotter; this is the compression or local expansion-history contribution, distinct from the clock-rate effect on the escaping radiation.

Multiplying the two first-order factors, and cancelling the common subsequent background redshift, givesFor the overdense region produced by adiabatic compression, $\Delta R<0$, so the final term is positive. It competes with the negative gravitational redshift term. Their sum is the intrinsic-plus-endpoint part of the Sachs-Wolfe combination; computing a full observed Cosmic microwave background anisotropy can also require Doppler CMB anisotropy and the Integrated Sachs-Wolfe effect.^[ref]

A stationary black hole behaves mechanically like an equilibrium thermodynamic system. Its horizon area A, asymptotic energy E, angular momentum J and electric charge Q are linked by the laws of black-hole mechanics. Let ${\chi }^a=t^a+{\Omega }_H{\varphi }^a$ generate its Killing horizon, with the time translation normalized at infinity; its surface gravity is defined by ${\chi }^b{\nabla }_b{\chi }^a=\kappa {\chi }^a$ on that horizon. We use $c=1$ and keep G explicit. The horizon electrostatic potential ${\Phi }_H$ is measured relative to infinity.

The Zeroth law of black-hole mechanics states that kappa is constant on each connected stationary horizon, under the Einstein equations and suitable energy/regularity assumptions. It parallels the zeroth law of thermodynamics, which makes temperature uniform in equilibrium. Kappa need not be constant in a dynamical geometry. For an asymptotically flat static hole, local observers see gravitationally redshifted temperatures; the equilibrium quantity comparable to kappa is the temperature referenced to infinity.

The First law of black-hole mechanics relates nearby stationary solutions:The same leading relation describes a small physical process which perturbs a horizon and subsequently settles to equilibrium, with the appropriate hypotheses on fluxes and final state. The work terms correspond to rotational and electromagnetic work. The structural analogy with the first law of thermodynamics, $dE=TdS+\text{work}$, identifies area as an entropy-like variable and surface gravity as a temperature-like variable. Classically this fixes their product in the first law, rather than their separate normalizations.

The second law of black-hole mechanics, or black-hole area theorem, states that future event-horizon area does not decrease classically. Its essential geometric ingredient is the Raychaudhuri equation for affinely parametrized null generators. In four dimensions a hypersurface-orthogonal congruence obeysThe null energy condition and Einstein equations make the last term nonpositive. If the expansion became negative, the inequality would focus the generators to a conjugate point in finite affine parameter. Under future completeness and predictability assumptions, that would conflict with their remaining on an achronal event horizon. Thus $\vartheta \ge 0$, and the area element, satisfying $d\log dA/d\lambda =\vartheta$, cannot decrease. For a merger the final area is at least the sum of the initial component areas. These assumptions matter: the theorem is a classical statement, not a prohibition of quantum evaporation.

The third law of black-hole mechanics is an unattainability statement: a nonextremal hole cannot be driven to zero surface gravity in a finite physical process satisfying the usual bounded-stress and regularity conditions. This parallels the unattainability form of the third law of thermodynamics. Extremal stationary solutions exist, so the law concerns how an initially nonextremal hole approaches them. The stronger entropy form of Nernst's theorem does not transfer unchanged: extremal holes can have finite area and entropy, depending on their conserved charges.

Quantum field theory turns the mechanical analogy into a physical thermodynamic identification. Hawking radiation is thermal at infinity to leading semiclassical order, withThese give $T_H\delta S_{\mathrm{BH}}=\kappa \delta A/(8\pi G)$ exactly, so the mechanical first law becomes the thermodynamic first law with Bekenstein-Hawking entropy. Scattering through the exterior adds greybody factors to the emitted spectrum but leaves this temperature identification intact. Equilibrium with an external bath and isolated evaporation are different physical states of the field.

During black-hole evaporation, the area can decrease because the semiclassical stress tensor need not satisfy the classical null energy condition. The proposed generalized second law applies instead to $S_{\mathrm{gen}}=S_{\mathrm{BH}}+S_{\mathrm{outside}}$, asserting nondecrease of the combined entropy. A Schwarzschild hole illustrates an unusual thermodynamic property: $T_H\propto E^{-1}$, so its heat capacity is negative. It becomes hotter as it loses energy and is unstable in a simple canonical heat bath, while entropy and energy laws still describe an isolated system. The quantum temperature, area entropy and generalized second law supply the correspondence; the mechanical laws alone do not establish a thermal emission spectrum or a microscopic entropy count.

Use $\mathcal{R}$ for the stellar gas constant, reserving $R$ for the stellar radius. The specific internal energy of a monatomic gas obeying the ideal gas law is $e=P/[(\gamma -1)\rho ]=3P/(2\rho )$. At fixed enclosed mass, the first law of thermodynamics gives the heat available to the outgoing luminosity as $ϵ=-{\partial }_{t}e-P{\partial }_t(1/\rho )$ when there is no stellar nuclear fusion heating. ThereforeThe second coefficient includes both the mass density dependence of $e$ and pressure work. This is gravitational energy generation in a homologously contracting ideal-gas star.

Write the characteristic scales asThen $\rho ={\rho }_{0}b$, $P=P_{0}p$, $T=T_{0}p/b$, $m=Mq$ and $L_r=Ll$. Since $d/dr=R^{-1}d/dx$, direct substitution in the stellar hydrostatic equation and mass conservation givesFor radiative diffusion in a star, dividing the temperature equation by $T_0/R$ givesIn particular the fourth power here is of $\mathcal{R}$, not of the radius.

Because $q(x)$ is fixed and $M$ is constant, a fixed mass label has fixed $x$. Thus stellar homology gives ${\partial }_t\rho =-3\rho \dot{R}/R$ and ${\partial }_{t}P=-4P\dot{R}/R$. The heating rate reduces to $ϵ=-3P\dot{R}/(2\rho R)$, positive during Kelvin-Helmholtz contraction. Substitution in $d{L}_r/dr=4\pi r^2\rho ϵ$ now yieldsThe dimensionless profiles fix $D$ and $E$. In a common stellar homology family with fixed stellar composition and opacity normalization, the expression for $D$ impliesIt is independent of radius and constant in time for a particular fixed-mass star. The formula for $E$ gives $\dot{R}=-2EL{R}^2/(3G{M}^2)$, which integrates toWhen the initial-radius term is negligible,This is the Kelvin-Helmholtz contraction time law in the prescribed stellar homology model. Infinite initial radius is a limiting idealization; a finite initial radius retains the first term.

At main sequence arrival, CNO cycle heating with fixed stellar composition has luminosityThe integral is dimensionless and fixed within the stellar homology family. The constant-opacity radiative scaling still gives $L\propto M^3$. Equating the stellar nuclear fusion and transported luminosities givesFinally the Kelvin-Helmholtz contraction law evaluated at $R_{\mathrm{MS}}$ gives $t_{\mathrm{MS}}\propto M^2/(L{R}_{\mathrm{MS}})\propto M^{-34/19}$. In a coeval star cluster, the mass just arriving at the main sequence consequently satisfiesThese results comprise constant-opacity homologous contraction to CNO ignition; their mass exponents compare fixed-composition models with common dimensionless profiles.

Take any fixed material subvolume $B_0$ of the reference configuration, with outward normal vector $N$, current particle velocity $v_i={\dot{x}}_i$, and positive temperature $\theta$. The first law of thermodynamics, including kinetic energy, isThe heat-flux sign is negative because $N_I{q}_I^0$ is outward. The Second law of thermodynamics supplies the corresponding integral Lagrangian entropy inequality:The temperature in each supply or flux term is the local temperature, not a common temperature pulled outside the integrals.

Apply the divergence theorem and use the reference momentum conservation equation ${\rho }_0{\dot{v}}_i=P_{Ii,I}+{\rho }_0{g}_i$. In the total-energy equation, the kinetic derivative cancels the body-force work and the $v_i{P}_{Ii,I}$ term. Since $v_{i,I}={\dot{F}}_{iI}$, the remaining local Lagrangian internal energy balance and Lagrangian entropy inequality areThey hold pointwise because $B_0$ was arbitrary. In particular,

The specific Helmholtz free energy is $\psi =u-\theta \eta$. Substitution yields its balanceand elimination of the heat supply yields the Clausius-Duhem inequalityFor $\psi =\psi (F,\theta ,{\xi }_r)$, expand the derivative to obtainThe usual Coleman-Noll procedure admits independent reversible choices of $\dot{F}$ and $\dot{\theta }$ at a fixed state. Their coefficients must vanish, since either sign and arbitrary magnitude would otherwise violate the inequality. ThusInternal-variable evolution and heat conduction must satisfy this remaining inequality. Substituting the constitutive relations back into the energy balance givesThe last printed equation is missing a factor ${\rho }_0$ on its internal-variable term. The PDF defines $f_r=-{\psi }_{,{\xi }_r}$ with $\psi$ per unit mass, so the expression above is the dimensionally consistent consequence. Equivalently define the volumetric thermodynamic force conjugate to an internal variable ${\hat{f}}_r=-{\rho }_0{\psi }_{,{\xi }_r}$; the term is then ${\hat{f}}_r{\dot{\xi }}_r$ with no extra density. The two conventions must not be mixed.

For thermoelasticity with a temperature-dependent constraint, admissible rates obeyThe reversible coefficient functional need now vanish only on this tangent hyperplane. Hence it can be a scalar multiple $q$ of the constraint normal:ThereforeThe reaction terms cancel on admissible rates, leaving the same internal-variable/conduction inequality and entropy heating equation. The multiplier $q$ is determined by equilibrium and boundary conditions rather than by the free energy alone.

For linear thermoelasticity with prescribed thermal volume, write $F=I+H$, $H_{ij}=u_{i,j}$, and $\vartheta =\theta -{\theta }_0$. Hold the internal variables fixed in this reversible linearization, and evaluate all derivatives below at $(I,{\theta }_0)$ and zero reaction. The stress-free reference state has ${\psi }_{,F}=0$. Since $\partial \det F/\partial F=I$ there, first-order nominal and Cauchy stress tensors agree. Their constitutive laws becomewhere equality of mixed free-energy derivatives givesThe linearized constraint is $u_{i,i}=h^{\prime }({\theta }_0)\vartheta$. Thus volume is thermally prescribed; strict temperature-independent incompressibility corresponds to $h^{\prime }=0$. The coefficient $C_e$ is an entropy slope; the corresponding volumetric heat capacity at fixed deformation is ${\theta }_0{C}_e$.

Together with ${\rho }_0{\ddot{u}}_i={\sigma }_{ji,j}+{\rho }_0{g}_i$, the first-order thermal balance is ${\theta }_0(C_e\dot{\vartheta }-{\beta }_{ji}{\dot{u}}_{i,j}+h^{\prime }({\theta }_0)\dot{q})={\rho }_{0}r-q_{I,I}^0$. A heat-flux constitutive law completes the evolution problem. The paired signs of the coupling tensor follow from the same free energy, rather than from separately postulated mechanical and thermal equations.

A Lagrangian submanifold of a $2n$-dimensional symplectic manifold $(M,\omega )$ is an $n$-dimensional submanifold $L$ such that the pullback of $\omega$ to $L$ is zero. Equivalently, each tangent space $T_{x}L$ is its own symplectic orthogonal. The dimension condition makes it maximally isotropic: in a nondegenerate alternating vector space no isotropic subspace has dimension greater than $n$.

In geometric quantization, a line bundle with a connection on a vector bundle whose curvature is proportional to $\omega /\hslash$ provides the initial quantum-state space. A real polarization foliates phase space locally by Lagrangian leaves; requiring sections to be covariantly constant along these leaves reduces their independent variables by half. The curvature restricts to zero on a Lagrangian leaf, allowing local flat sections. Globally its holonomy can impose the familiar Bohr-Sommerfeld quantization condition. Thus Lagrangian submanifolds describe choices of quantum configuration data and the support of semiclassical states, rather than requiring simultaneous sharp values of all position and momentum coordinates.

For the thermodynamic application, regard $(S,V,N)$ as base coordinates $q^i$ and $(T,-P,\mu )$ as fiber coordinates $p_i$ on $T^{\ast }{\mathbb{R}}^3$. The canonical one-form on a cotangent bundle and the position-first symplectic form areThe negative sign in the pressure coordinate is essential. A smooth potential $M(S,V,N)$ determines the embeddingwhere the last three coordinates are ordered as $(T,-P,\mu )$: explicitly $T=M_S$, $P=-M_V$, and $\mu =M_N$, so the second fiber coordinate is $-P=M_V$. Accordingly, in the actual $p_i$ coordinates the graph isProjection to the base is the identity, so this is an embedded three-dimensional submanifold, even if the Hessian of $M$ is degenerate. The first law of thermodynamics gives ${\iota }_M^{\ast }\theta =dM$. HenceIt has half the dimension of the ambient cotangent bundle and is therefore Lagrangian. This proves the thermodynamic potential as a Lagrangian graph construction for every smooth choice of $M$, locally on any domain on which the potential is defined. No invertibility, extensivity or stability assumption is needed for this geometric conclusion.

In components the same vanishing follows from equality of mixed derivatives. The thermodynamic conjugate variables satisfyThese cancel the three coefficients of the pulled-back two-form. They are the integrability relations behind the local thermodynamic potential, and demonstrate directly why the first-law graph is Lagrangian.

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.

Write the photon energy density as $ϵ=a_R{T}^4$, where $a_R=4\sigma /c$ by the Stefan–Boltzmann law. The photon gas has pressure $P=ϵ/3$ and zero photon chemical potential. The first law of thermodynamics therefore givesDividing by $T$ and integrating gives the entropy density of a photon gas:The integration constant is fixed to zero at zero temperature, consistently with the third law of thermodynamics.

The inequality $\Gamma \gg H$ says interactions act much faster than the expansion time, maintaining local thermal equilibrium. For adiabatic reversible expansion with no additional source of entropy, cosmological entropy conservation fixes the entropy in a comoving volume. Since that volume scales as $a^3$, and the relativistic degrees of freedom are fixed here, $T^3{a}^3$ is constant. ThusDuring radiation domination, the Friedmann equation gives $H^2\propto \rho \propto T^4$, and hence $\boxed{H\propto T^2}$. Fast interactions alone would not exclude an additional entropy-producing process; changing relativistic species likewise changes the simple temperature scaling.

Take a fixed comoving volume, whose physical volume is $V=V_{\ast }{a}^3$. For adiabatic cosmic expansion, the first law of thermodynamics gives $d(\rho V)=-PdV$. ThereforeSpatial curvature does not change the $a^3$ scaling of a fixed comoving volume. With $C=8\pi G{\rho }_{\ast }/3$ and $d=1+3w$, the Friedmann equation becomesDifferentiating this expression, rather than dividing by ${\Omega }_k$ at a fixed point, yields the cosmological density parameter flowThis holds on an expanding branch with $H\ne 0$. The spatially flat solution ${\Omega }_k=0$ is a fixed point. For $|{\Omega }_k|\ll 1$, ${\Omega }_k\propto a^{1+3w}$. Consequently curvature deviations grow as $a^2$ during radiation domination and as $a$ during matter domination. A small present curvature therefore requires a much smaller initial deviation: this is the flatness problem of a decelerating Big Bang. The issue applies to either sign of curvature, not just an open universe. Conversely, accelerated expansion with $w<-1/3$ suppresses small deviations, which is the inflationary solution of the flatness problem.

The stellar adiabatic exponents are fixed-composition, constant-specific entropy derivatives:The chain rule gives ${\Gamma }_1({\Gamma }_2-1)/{\Gamma }_2={\Gamma }_3-1$. Define the pressure derivatives ${\chi }_{\rho }=(\partial \log P/\partial \log \rho {)}_T$ and ${\chi }_T=(\partial \log P/\partial \log T{)}_{\rho }$. The first law of thermodynamics, with $du=Tds+Pd\rho /{\rho }^2$, givesCombining this with $d\log P={\chi }_{\rho }d\log \rho +{\chi }_{T}d\log T$ on an adiabat givesThus the specific-heat ratio is not generally equal to the three stellar adiabatic exponents.