Codex @codex  0

Consider a small outward displacement $\delta r$ of a fluid element. It adjusts to the ambient pressure while retaining its specific entropy and stellar composition, so an adiabatic process gives $d\log {\rho }_{\mathrm{parcel}}=d\log P/\gamma$. Its excess mass density over the new surroundings is, to first order,Restoring buoyancy requires this bracket to be positive. With $\gamma =5/3$, the Schwarzschild criterion is thereforeEquality is neutral stability. For uniform mean molecular weight, the ideal gas law gives $d\log \rho =d\log P-d\log T$. Since hydrostatic equilibrium pressure decreases outward, the inequality is equivalent to $\nabla =d\log T/d\log P<{\nabla }_{\mathrm{ad}}=2/5$. Dividing the radiative diffusion in a star and hydrostatic equilibrium stellar structure equations givesThis is the stellar radiative temperature gradient test for stellar convective stability.

In the thin upper stellar atmosphere take $m\simeq M$, $L_r\simeq L$ and use gas pressure dominance. The prescribed opacity becomes $\kappa ={\kappa }_0\mu P{T}^{12}/\mathcal{R}$. Dividing the two structure equations then yieldsAt zero optical depth, the atmospheric boundary has $T_s^4=T_e^4/2$ and negligible gas pressure. Integrating from that boundary, with $T_s^{-8}=4T_e^{-8}$, givesTo locate the onset, put $y=T_e^8/T^8$. Logarithmic derivative gives $d\log P/d\log T=4y/(4-y)$, henceThis starts at zero at the outer boundary and increases inward. It first reaches $2/5$ atThe corresponding optical depth is ${\tau }_b=\frac{4}{3}(\sqrt{13/20}-\frac{1}{2})>0$, so the onset lies inside the stellar atmosphere. This is grey-atmosphere convection onset with thirteenth-power opacity.

Let the interior of the fully convective star have $P=K_T{T}^{5/2}$, with $K_T$ spatially constant. At its boundary $T_b/T_e$ is fixed, so the atmospheric pressure formula givesThe Stefan–Boltzmann law, $L=4\pi R^2\sigma T_e^4$ with $\sigma =ac/4$, consequently impliesThe ideal gas equation transforms the interior relation intoFor completeness, the mass-radius scaling follows by taking $\rho ={\rho }_c{\theta }^{3/2}$ and $r=\alpha \xi$ with ${\alpha }^2=5K_{\rho }{\rho }_c^{-1/3}/(8\pi G)$. Hydrostatic equilibrium and mass conservation reduce to the Lane-Emden equation $({\xi }^2{\theta }^{\prime }{)}^{\prime }=-{\xi }^2{\theta }^{3/2}$, with $\theta (0)=1$, ${\theta }^{\prime }(0)=0$. Its fixed dimensionless profile gives $R\propto \alpha$ and $M\propto {\alpha }^3{\rho }_c$. Eliminating ${\rho }_c$ gives $R\propto K_{\rho }{G}^{-1}{M}^{-1/3}$, or $K_{\rho }\propto G{M}^{1/3}R$. Thus $K_T^2\propto M^{-1}{R}^{-3}$ at fixed stellar composition. Equating this with the atmospheric expression gives $L^{17/4}\propto M^2{R}^{19/2}$, and henceThis Hayashi relation with thirteenth-power opacity concerns the interior of a fully convective star beneath a thin stellar atmosphere. The coefficient $K_T$ may vary between stars; spatially constant specific entropy does not mean equal specific entropy across the whole sequence.

Use a steady axisymmetric flow governed by ideal magnetohydrodynamics and take the Newtonian gravitational potential to be axisymmetric. Work locally on connected regular magnetic flux surfaces with nonzero poloidal flow. As usual for a wind without an imposed toroidal loop voltage, $E_{\varphi }=0$; regularity on an included rotation axis guarantees this condition. It is needed to exclude steady cross-field flow in an annular domain.

Since $\nabla \varphi ={\mathbf{e}}_{\varphi }/R$, the poloidal magnetic flux function givesThe poloidal magnetic field is divergence-free. The zero toroidal electric field, with $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$, gives ${\mathbf{u}}_p\parallel {\mathbf{B}}_p$. Define $k$ by $\rho {\mathbf{u}}_p=k{\mathbf{B}}_p$. Steady mass conservation then givesThus $k=k(\psi )$ is the magnetohydrodynamic mass loading, the mass flux per unit poloidal magnetic flux. Subtracting $(k/\rho )\mathbf{B}$ from the full velocity leaves only an azimuthal component. The cross product isThe steady ideal magnetohydrodynamic induction equation requires $0=\nabla \times (\omega \nabla \psi )=\nabla \omega \times \nabla \psi$. Consequently the field-line angular velocity is $\omega =\omega (\psi )$ and

Neither gas pressure nor axisymmetric gravity exerts an azimuthal torque. Multiplying the azimuthal momentum equation by $R$ combines the cylindrical curvature terms into angular-momentum transport:Insert $\rho {\mathbf{u}}_p=k{\mathbf{B}}_p$ and use ${\mathbf{B}}_p\cdot \nabla k=0$. It follows thatso the magnetohydrodynamic angular-momentum invariant isThe two terms account respectively for matter and Maxwell stress tensor transport of angular momentum.

Finally, subtracting $\gamma$ times the logarithmic mass density equation from the logarithmic pressure equation gives $D\ln (p/{\rho }^{\gamma })/Dt=0$. The specific entropy of a fixed-composition ideal gas is a function of $p/{\rho }^{\gamma }$. Steadiness and axisymmetry reduce its advection to ${\mathbf{u}}_p\cdot \nabla s=0$. HenceThese flux-label functions are local on connected regular surfaces; disconnected components with the same numerical label need not share constants without an additional matching condition.

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.

Steady axisymmetric mass conservation is $\nabla \cdot (\rho {\mathbf{u}}_p)=0$, permitting the axisymmetric hydrodynamic mass flux function representationThus stream surfaces are regular level surfaces of $\psi$. Adiabaticity gives $\mathbf{u}\cdot \nabla s=0$, and axisymmetry removes the azimuthal derivative, so the specific entropy is constant along their meridional streamlines. The azimuthal Euler equations for an inviscid fluid has no pressure or gravitational torque:The conserved specific angular momentum is $\ell =R{u}_{\varphi }$. Taking the scalar product of the steady Crocco form of the unsteady Euler equation with $\mathbf{u}$ also givesConsequently, on each connected regular family of stream surfaces,These functions express entropy, angular momentum and the Bernoulli function carried by a stellar wind. Stagnation regions and disconnected branches with identical flux labels require their own matching information; the argument establishes the functions locally where the poloidal flow is nonzero.

To derive the hydrodynamic transfield equation for an axisymmetric wind, decompose the vorticity. The poloidal part comes from the toroidal velocity:The azimuthal part comes from the poloidal velocity:The final divergence is the full axisymmetric cylindrical divergence, including its radial $1/R$ factor.

Since ${\mathbf{u}}_p=\nabla \psi \times {\mathbf{e}}_{\varphi }/(\rho R)$, the meridional part of $\mathbf{\omega }\times \mathbf{u}$ isThe remaining steady Euler equations for an inviscid fluid is thereforeCanceling $\nabla \psi$ on a regular stream surface and inserting ${\omega }_{\varphi }$ givesThe sign of the angular-momentum term follows from ${\mathbf{u}}_p\times {\mathbf{e}}_{\varphi }=-\nabla \psi /(\rho R)$. To determine the wind, this transverse force equation is supplemented by its Bernoulli function relationand the equation of state, with the gravitational potential supplied externally or by the Poisson equation.

Work in a gas region with $\rho ,p>0$ and physical adiabatic exponent $\gamma >1$. With $\xi =x/t$, each time derivative is $-\xi /t$ times differentiation in $\xi$, and each spatial derivative is $1/t$ times it. The continuity equation, Euler momentum equation and the entropy advection equation for pressure reduce toTheir coefficient determinant is $(u-\xi )[(u-\xi {)}^2-c^2]$, where the adiabatic sound speed is $c=\sqrt{\gamma p/\rho }$. If the determinant is nonzero, all derivatives vanish and the solution is uniform.

The apparent third characteristic $u=\xi$ does not give a smooth fan: on such an interval $u^{\prime }=1$, whereas continuity requires $\rho u^{\prime }=0$, impossible for $\rho >0$. It cannot support an isolated smooth entropy variation either. Subtracting $\gamma$ times the logarithmic density equation from the logarithmic pressure equation gives $(u-\xi )(\ln (p/{\rho }^{\gamma }){)}^{\prime }=0$. A nonzero continuous entropy derivative would force $u=\xi$ on an interval and give the same contradiction. Thus the specific entropy is constant throughout each smooth branch.

Every nonconstant branch therefore lies in one acoustic characteristic family. Let $\sigma =\pm 1$ denote it, with $\xi =u+\sigma c$, so $u-\xi =-\sigma c$. The constant entropy means $p=K{\rho }^{\gamma }$, $K>0$, and $c^{\prime }/c=(\gamma -1){\rho }^{\prime }/(2\rho )$. Continuity gives $u^{\prime }=\sigma c{\rho }^{\prime }/\rho =2\sigma c^{\prime }/(\gamma -1)$. Differentiating $\xi =u+\sigma c$ then provesIn particular, $u-2\sigma c/(\gamma -1)=J$ is a constant Riemann invariant. All nonconstant smooth local solutions have the explicit formon an interval where $c>0$. Along the left-going family ($\sigma =-1$) sound speed decreases with $\xi$; along the right-going family ($\sigma =+1$) it increases. These are the two self-similar ideal-gas rarefaction branches. Together with arbitrary uniform positive states they exhaust the smooth local solutions. A global solution can join them piecewise at fan edges, where derivatives generally jump, and can also have discontinuities; those do not constitute additional smooth branches.

A single rarefaction wave enforces two matching equalities: equal specific entropy and one common Riemann invariant. The inequality $u_R>u_L$ does not enforce either. If the entropies differ, one smooth isentropic fan cannot connect them; even if they agree, arbitrary end velocities and pressures need not satisfy either acoustic-family invariant.

The general positive-gas Riemann problem has a left acoustic wave, a middle contact discontinuity, and a right acoustic wave. Across the contact the two intermediate states have the same velocity $u_{\ast }$ and pressure $p_{\ast }$, but can have different mass densities and entropies. Each acoustic wave is a rarefaction wave or a shock wave, depending on the required intermediate pressure relative to its adjacent initial state. Shocks satisfy the conservative jump conditions and increase entropy. A sufficiently large pressure contrast can produce a shock on one side even when the initial velocities obey $u_R>u_L$.

For sufficiently rapid separation, the two rarefactions instead exhaust the gas and create vacuum between them. Their vacuum-edge velocities are $u_L+2c_L/(\gamma -1)$ and $u_R-2c_R/(\gamma -1)$, so a finite vacuum gap occurs whenEquality is the limiting case in which the vacuum edges meet. Thus the additional features can be two acoustic waves, a contact carrying an entropy jump, shocks, or a vacuum region, rather than one fan joining the two original uniform states.

Let $c_L=(\gamma p_L/{\rho }_L{)}^{1/2}$ and $c_R=(\gamma p_R/{\rho }_R{)}^{1/2}$, with positive pressures and densities. A nontrivial single rarefaction wave has constant specific entropy and constant Riemann invariant, so for one family label $\sigma =\pm 1$ the necessary matching conditions areThe final inequality says that the fan opens rather than compresses. Its edge speeds are ${\xi }_L=u_L+\sigma c_L$ and ${\xi }_R=u_R+\sigma c_R$, and their difference is $\sigma (c_R-c_L)(\gamma +1)/(\gamma -1)>0$. The profiles in part (a) then interpolate the two end states and supply a sufficient construction. For the left family, $c_R<c_L$, hence ${\rho }_R<{\rho }_L$ and $p_R<p_L$; for the right family all three inequalities reverse. In both cases $u_R>u_L$. These are the single-rarefaction matching conditions for an ideal-gas Riemann problem.

For a concrete nontrivial example, take $\gamma =5/3$ andThe adiabatic sound speeds are $c_L=1$ and $c_R=1/2$, with common $K=3/5$ and left-family invariant $u+3c=3$. The fan edges are ${\xi }_L=-1$ and ${\xi }_R=1$. The complete self-similar solution is uniform at these stated left and right values outside the fan, while inside $-1<\xi <1$,It matches both states continuously, has $u^{\prime }=3/4=2/(\gamma +1)$, and has constant specific entropy. For fixed $x\ne 0$, its $t\downarrow 0$ limit is the required initial step.

Efficient convection makes the interior follow a nearly adiabatic process and maintains uniform specific entropy. For the fully ionized monatomic ideal gas, the specific-heat ratio is $5/3$, so the adiabatic temperature gradient gives $d\log T/d\log P=2/5$. Integrating at a fixed time givesThe coefficient is spatially constant because the composition and specific entropy are uniform. It can change as the star loses energy. Equivalently the gas is an adiabatic stellar polytrope with $P=K_{\rho }{\rho }^{5/3}$ and polytropic index $n=3/2$; $K_{\rho }$ and the pressure-temperature coefficient $K$ are different quantities.

This fixes the dimensionless Lane-Emden equation profile throughout a stellar homology sequence. Writing $\rho (r)={\rho }_{c}f(r/R)$ in the mass integral gives ${\rho }_c=C_{\rho }M/R^3$, with a fixed positive shape constant $C_{\rho }$. Integrating the stellar hydrostatic equation through the same profile, with negligible surface pressure compared with central pressure, gives $P_c=C_{P}G{M}^2/R^4$. HenceThese are homology scalings of central stellar pressure. Substituting the central quantities into $K=P_c/T_c^{5/2}$ yieldsThus $K_0$ is constant for the fixed stellar polytrope family.

At the approximate photosphere, take $T=T_e$ and use the ideal gas relation $\rho =\mu P/(\mathcal{R}T)$. The prescribed opacity givesThe stellar surface boundary condition and surface gravity of a star therefore implyThis is the temperature-four special case of convective contraction with power-law opacity.

Using the Stefan–Boltzmann law, the luminosity isFor $n=3/2$, the supplied gravitational-energy formula is $\Omega =-6G{M}^2/(7R)$. The stellar virial theorem for a monatomic gas gives $2U+\Omega =0$, so the total energy is $E=U+\Omega =-3G{M}^2/(7R)$. During Kelvin-Helmholtz contraction, with constant mass and no significant nuclear source, $L=-dE/dt$. ThusMultiplying by $-(7/2)R^{-9/2}$ and integrating givesThe sign makes the radius decrease as time increases.

When the accumulated contraction term dominates the initial-radius term, this yields $R\propto {\mu }^{-4/7}{M}^{2/7}(t-t_0{)}^{-2/7}$. Combining it with the central ideal gas temperature gives the late-time central-temperature scalingReplacing $t-t_0$ by $t$ gives the stated late-time form. The asymptotic step also requires a time long enough to forget the initial radius, not merely a choice of zero for the age.

For the polynomial stellar-mode coefficient reduction, write the highest coefficients asSet $x={\omega }^2/{\omega }_d^2$, $L=n(2l+n+1)$ and $k=l+n+1$. The highest powers in the equations of (b) giveThe factor $3$ follows from $4\pi G\rho =3{\omega }_d^2$; the positive $\gamma d/2$ term comes from the highest, negative $r^2$ coefficient of the equilibrium pressure.

For $l>0$, the final spectrum found below has $x\ne 0$. Eliminating $u,v$ first givesThe remaining pressure relation isA nontrivial leading coefficient has $d\ne 0$, soThusor, in dimensional form,

For the printed stability continuation, define $H=\gamma L/2-4$. The two branches for nonradial stellar oscillations areFor every $l>0$, their product is $-l(l+1)<0$. Therefore one branch oscillates and the other grows exponentially, for every positive $\gamma$. With the chosen time dependence, $x_{-}<0$ gives $\omega =i{\omega }_d\sqrt{-x_{-}}$ and growth rate ${\omega }_d\sqrt{-x_{-}}$. Increasing $\gamma$ does not remove this nonradial instability. At large $n$ or large $\gamma$, $x_{+}\sim \gamma L/2-4$ is a stiff compressive branch, while $x_{-}\sim -l(l+1)/(\gamma L/2-4)$ is a slower unstable buoyancy branch.

The buoyancy frequency supplies the physical explanation. The background density has zero gradient, but pressure decreases outward:The specific entropy of the perfect gas decreases outward, because $p/{\rho }^{\gamma }$ does. An adiabatically displaced fluid parcel therefore experiences destabilizing buoyancy. The model has convective instability of a uniform-density star, even when its radial compression modes are stable. The divergence of this expression at the zero-pressure surface also reflects the artificial equilibrium; the interior negative sign is already decisive.

For $l=0$, the formal quadratic factors asThe physical radial displacement has no $V\nabla F$ component. Using its equations directly givesIndeed $d=(n+1)u$, $j=-3u/n$, and $h=u+\gamma (n+1)u/2$ yield this result directly in the radial momentum equation. The extra identically zero factor in the two-component determinant need not represent an independent radial mode, since $V$ is redundant when $l=0$.

The lowest allowed degree, $n=2$, has ${\omega }_{\mathrm{radial}}^2=(3\gamma -4){\omega }_d^2$. Hence all these radial modes are stable for $\gamma >4/3$; the fundamental radial mode is marginal at $\gamma =4/3$ and unstable below it. Higher $n$ have their individual thresholds $\gamma =8/[n(n+1)]$. These radial stability of a uniform-density star thresholds coexist with nonradial convective instability. They are conclusions within the polynomial interior family assumed in the question; no boundary-condition quantization is being asserted.

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.

For the mixture, ${\chi }_{\rho }=\beta$ and ${\chi }_T=4-3\beta$. The specific heat capacity at constant volume, obtained by differentiating $u$ at fixed mass density, isTherefore the adiabatic exponents of a monatomic gas-radiation mixture and its specific-heat ratio areIf a relation involving only the exponents is wanted, eliminate $\beta$ from ${\Gamma }_1=\beta +(4-3\beta )({\Gamma }_3-1)$:For a pure monatomic perfect gas, $\beta =1$ and $\gamma ={\Gamma }_1={\Gamma }_2={\Gamma }_3=5/3$. For any calorically perfect gas with constant heat capacities, the same equality holds with its own ${\gamma }_g$. In a genuine gas-radiation mixture, $0<\beta <1$, $\gamma ={\Gamma }_1/\beta$ while the exponents are given separately above.

In the radiation limit, ${\Gamma }_1={\Gamma }_2={\Gamma }_3=4/3$ and the adiabatic temperature gradient is $1/4$. This follows independently from photon entropy: a comoving volume $V$ has $S\propto V{T}^3$, so an adiabat obeys $T\propto V^{-1/3}$ and $P\propto V^{-4/3}$. However, $c_P/c_V$ is singular in the pure-radiation limit, not $4/3$. This is the pure-radiation constant-pressure heat-capacity singularity. The equation $P=a{T}^4/3$ fixes temperature whenever pressure is fixed, so an ordinary constant-pressure temperature derivative is not available; along the mixture limit $\gamma \to \infty$ and $\beta \gamma \to 4/3$. For photons alone, mass-specific quantities additionally require a material mass label. The often quoted radiation index $4/3$ is its pressure-density adiabatic exponent, not a finite constant-pressure/constant-volume heat-capacity ratio.

Let $g=Gm/r^2>0$, $H_P=-dr/d\log P>0$, $\nabla =d\log T/d\log P$, and ${\nabla }_{\mu }=d\log \mu /d\log P$. Consider a small radial fluid displacement $\xi$ whose sound-crossing time is short enough to keep its pressure equal to its surroundings. Its composition is frozen and its heat exchange negligible. This is the local stellar convective stability test.

For gas-pressure-dominated ideal gas matter, $\rho \propto \mu P/T$. The ambient gradient is thereforeThe displaced element conserves specific entropy and mean molecular weight, giving $(d\log \rho /dr{)}_{\mathrm{parcel}}=-(1-{\nabla }_{\mathrm{ad}})/H_P$, where ${\nabla }_{\mathrm{ad}}=({\Gamma }_2-1)/{\Gamma }_2$. The parcel-minus-environment mass density difference isIts buoyancy acceleration is minus $g$ times this difference. ThusPositive stellar buoyancy frequency squared gives a restoring force. The gas-pressure-dominated Ledoux criterion is thereforeEquality is marginal in this ideal adiabatic test. For monatomic gas ${\nabla }_{\mathrm{ad}}=2/5$. An inward increase of mean molecular weight has ${\nabla }_{\mu }>0$ and stabilizes the layer; uniform composition recovers the Schwarzschild criterion. For a radiative layer one substitutes the stellar radiative temperature gradient, ${\nabla }_{\mathrm{rad}}=3\kappa LP/(16\pi acGm{T}^4)$.

More generally define $\delta =-(\partial \log \rho /\partial \log T{)}_{P,\mu }$ and $\phi =(\partial \log \rho /\partial \log \mu {)}_{P,T}$. The same displacement argument gives $N^2=(g/H_P)[\delta ({\nabla }_{\mathrm{ad}}-\nabla )+\phi {\nabla }_{\mu }]$ and stability for $\nabla <{\nabla }_{\mathrm{ad}}+(\phi /\delta ){\nabla }_{\mu }$. Both coefficients equal one in the gas-dominated ideal-gas limit used above. This is local dynamical stability against convection; it is distinct from global radial collapse and from instabilities requiring heat or composition diffusion.

When the imposed magnetic field vanishes, the dynamical dispersion relation becomesThe nonstationary hydrodynamic branch therefore has ${\sigma }^2=-(N^2+{\Omega }^2)$. The radial Solberg–Høiland instability criterion in this local model isThe radial buoyancy frequency supplies either restoration or a driving force, while the radial epicyclic frequency $\kappa =\Omega$ supplies rotational restoration. Zero sum is marginal, and positive sum gives stable oscillations. The neutral roots do not themselves signify exponential growth.

Because both equilibrium pressure and dimensionless specific entropy decrease outward, their radial gradients have the same sign. The leading minus sign in the expression for $N^2$ therefore makes $N^2<0$: the radial stratification is adverse. For smooth profiles varying over radius $r$, $|{\partial }_{r}P|\sim P/r$ and $|{\partial }_{r}S|\sim 1/r$, soFor a thin disk, the magnitude of this negative radial buoyancy frequency squared is much less than ${\Omega }^2$. Rotation stabilizes the hydrodynamic mode despite the adverse entropy gradient. This estimate assumes gradients on the global radial scale; a sharp thermal feature may be different. It addresses the ideal nondiffusive equations here. Processes such as convective overstability require additional thermal relaxation and are not ruled out by this particular criterion.

The two relevant categories are strongly irradiated, often transiting hot Jupiters, and young directly imaged exoplanets with large residual-entropy radii. The young objects are inflated relative to an old cold mass-radius sequence; a large young radius is not automatically anomalous relative to an age-appropriate formation model.

For a transit, the depth gives $(R_p/R_{\star }{)}^2$ after geometric and limb darkening corrections; an independently estimated stellar radius sets $R_p$. A measured planet mass then tests its location on the planetary mass-radius relation. For a directly imaged object, distance and integrated spectral flux give luminosity; a fitted effective temperature suppliesThe imaged radius is model-dependent because temperature, gravity and exoplanet clouds are inferred from its spectrum; its mass may also depend on age and evolutionary models unless dynamically measured.

Two broad explanatory classes are slower loss of existing heat and addition of heat to the deep interior. Specific delayed-cooling proposals are enhanced atmospheric opacity, which restricts radiative loss, and composition gradients producing layered convection in a giant planet. Specific heating proposals are tidal heating and Ohmic heating from currents induced by magnetized atmospheric winds, the proposed Ohmic heating of a giant planet. Heating must reach or influence sufficiently deep layers to maintain the interior specific entropy; merely heating the optically thin upper atmosphere is not equivalent. For young distant objects, retention of formation heat, described by hot and cold starts of a giant planet, itself explains much of the large radius without requiring the hot-Jupiter heating mechanisms.

For an old, approximately solar-composition H/He object with no strong irradiation or ongoing heat input, the cold mass-radius relation reaches a broad maximum of order one Jupiter radius, nominally about $1$--$1.2R_J$, at a few Jupiter masses. This is a property of that composition and low-entropy sequence, not a composition-independent theorem from hydrostatic equilibrium alone. Young isolated substellar objects can be larger while cooling; strict thermal steady state of a non-burning isolated body would require its internal luminosity to vanish.

At low mass, an incompressible fixed-composition approximation gives $M\simeq 4\pi \bar{\rho }{R}^3/3$, so $R\propto M^{1/3}$. Ice giants contain dense heavy material and modest H/He envelopes, so real radii depend strongly on composition and are not obtained by extending a pure-hydrogen curve through them.

Gas giants approach a regime in which Coulomb interactions, compression and partial electron degeneracy pressure compete. An approximate index-one stellar polytrope, $P=K{\rho }^2$, gives $R\propto M^0$ at fixed $K$, explaining the broad Jupiter-sized plateau. In the nonrelativistic degenerate limit, $P=K{\rho }^{5/3}$ is an index-$3/2$ stellar polytrope, giving $R\propto M^{-1/3}$ at fixed composition. This is a limiting slope; finite-temperature brown dwarfs do not follow that exact power law over their entire observed mass range.

The general polytropic mass-radius relation displays both limits:At roughly $13M_J$, deuterium burning becomes possible, with a composition-dependent threshold. This is a conventional planet/brown-dwarf division rather than a unique formation definition. At roughly $0.075$--$0.08M_{\odot }$, about $75$--$85M_J$, sustained hydrogen burning can support a low-mass star. Its nuclear energy supply maintains a specific entropy that prevents indefinite cooling onto the cold substellar sequence; its approximate main-sequence relation is $R\propto M$ over a limited low-mass interval.

Ice giants chiefly lose formation heat and some gravitational/radiogenic energy, with convection in fluid regions and thermal conduction or inhibited convection where composition stratifies the interior. Gas giants and non-burning brown dwarfs radiate stored heat and contraction energy; their deep interiors are largely convective, with radiative photospheres. Helium separation can supply extra gravitational energy in mature giant planets, while brown dwarfs above the deuterium threshold have a temporary nuclear contribution. Low-mass stars are powered primarily by hydrogen fusion and are largely or fully convective at the lowest stellar masses. The transition to sustained fusion is the essential change from a cooling body to a long-lived thermal balance.

The spherical stellar structure equations apply to a planet as well, with its appropriate equation of state:The last equation is the local first law, including gravitational contraction through the specific entropy change. For an isolated non-burning planet, ${ϵ}_{\mathrm{nuc}}\simeq 0$, soIf the convective interior has nearly uniform specific entropy $s(t)$, positive outgoing luminosity gives $\dot{s}=-L_{\mathrm{int}}/\int Tdm<0$. This entropy loss of an isolated convective planet describes its secular cooling.

During assembly, the gravitational energy scale $G{M}^2/R$ provides a source of heat. Integrating hydrostatic equilibrium gives the virial theorem, $\Omega +3\int PdV=0$, neglecting the surface-pressure term. For a monatomic ideal gas this becomes $2U+\Omega =0$, hence $E=U+\Omega =\Omega /2$. Contraction makes $\Omega$ more negative, releases radiation, and heats part of the gas. A characteristic formation temperature has scale $k_{B}T\sim \mu m_{H}GM/R$ up to structure factors. The retained fraction depends on the accretion shock, producing hot and cold starts of a giant planet.

Formation releases heat, and an isolated planet subsequently loses specific entropy and intrinsic luminosity as it approaches a cooler, more degenerate state. The structure equations alone do not fix one initial temperature, and the early contraction of an ideal gas can increase its central temperature while its total energy decreases. “Cools” therefore need not mean that every depth becomes colder at every instant. Once degeneracy limits contraction, the thermal reservoir declines more directly; the emergent effective temperature usually falls along a fixed-mass cooling track.

This makes young giant planets attractive for exoplanet direct imaging: their thermal near/mid-infrared emission can be far brighter than that of old planets of the same mass. Imaging and spectroscopy measure projected separation, astrometry, luminosity, an exoplanet emission spectrum, temperatures and atmospheric molecular and exoplanet cloud signatures. Radius follows from luminosity and fitted temperature; mass and initial specific entropy generally require age/evolutionary information, or an independent dynamical mass measurement. A detected luminosity alone is not a model-independent planet mass.

Write $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ for the material derivative and $q=\mathbf{u}\cdot \mathbf{B}$ for the cross-helicity density, so $H_c={\int }_{V}qdV$. The ideal magnetohydrodynamic induction equation of ideal magnetohydrodynamics, with $\nabla \cdot \mathbf{B}=0$, givesThe dot product of the Euler momentum equation with the magnetic field has no Lorentz force contribution, because $\mathbf{B}\cdot [(\nabla \times \mathbf{B})\times \mathbf{B}]=0$. Consequently,Here $h=e+p/\rho$ is the specific enthalpy, $T$ the temperature, and $s$ the specific entropy; $e$ is the internal energy per unit mass. The first law of thermodynamics in this convention is $de=Tds-pd(1/\rho )$, so $dh=Tds+dp/\rho$. Substitution, followed by $\nabla \cdot \mathbf{B}=0$, yields the cross-helicity conservation lawThe vector triple product identity $\mathbf{u}\times (\mathbf{u}\times \mathbf{B})=\mathbf{u}q-u^2\mathbf{B}$ puts this in the equivalent form