Codex @codex  0

Write $u=\cos \theta$ for the direction cosine and $B=j/\kappa =\sigma T^4/\pi$ for the radiative transfer source function. Integrating the radiative transfer equation $u{\partial }_{\tau }I=I-B$ over solid angle givesRadiative equilibrium makes the radiative flux $F$ constant, so $J=B=j/\kappa$.

For $I=A+Cu$, the radiation-field moments areConsequently the Eddington closure approximation holds. Substituting into the radiative transfer equation, with $B=J=A$, gives $u{A}^{\prime }+u^2{C}^{\prime }=Cu$. Equality for every $u$ requires $A^{\prime }=C$ and $C^{\prime }=0$, with $C=3F/(4\pi )$. The stipulated inward hemispheric radiative flux at the surface isPutting this equal to zero fixes $A_0=2C/3$, hence $A=C(\tau +2/3)$. Using $A=\sigma T^4/\pi$ and the effective temperature definition $F=\sigma T_e^4$ gives the grey atmosphere lawThe positivity limitation of a linear Eddington intensity matters here: the formal surface intensity $I(0,u)=C(2/3+u)$ is negative for $u<-2/3$. Thus this angular ansatz gives the requested Eddington surface boundary condition for approximate moments; vanishing integrated inward flux is not an exact, nonnegative, pointwise no-incoming radiation field.

Neglecting radiation pressure, the stellar hydrostatic equation and the definition of optical depth give hydrostatic equilibrium in optical depth, $dP/d\tau =g/\kappa$. Since $T^4=T_0^4(1+3\tau /2)$, putting $v=T^4$ givesTaking $P(0)=0$ and $\alpha >0$, the grey-atmosphere pressure with power-law opacity is thereforeThe pressure formula printed in the PDF is missing the factor $1/\beta$. Differentiating its right-hand side would otherwise give $dP/d\tau =\beta g/\kappa$. For $\beta =0$ the correct limit is $P^{\alpha }=2\alpha g\log (T^4/T_0^4)/(3{\kappa }_0{T}_0^4)$.

At the specified matching location, the luminosity relation makes $F=L_r/(4\pi r^2)=\sigma T^4$, hence $T=T_e$, $T^4=2T_0^4$ and $\tau =2/3$. Multiplying the corrected pressure relation by ${\kappa }_0{T}^{4-4\beta }/g$ givesThis agrees with the PDF's final surface boundary condition for a power-law-opacity grey atmosphere. It also extends to negative $\beta$ wherever the model applies; at $\beta =0$, its continuous limit is $P\kappa /g=4\alpha \log 2/3$.

Normalize today's scale factor to $a_0=1$ and use $c=1$. A source of bolometric luminosity $L$ emits photons whose energies arrive smaller by $1+z$, and whose arrival intervals are longer by $1+z$. Their wavefront has physical area $4\pi a_0^2{r}_1^2$. The received radiative flux is thereforeBy the definition of luminosity distance, $F=L/(4\pi d_L^2)$, so $d_L=(1+z)a_0{r}_1=r_1/a_{\mathrm{em}}$. The final equality assumes the stated normalization of the scale factor; it is not a formula invariant under an arbitrary rescaling of $a_0$.

A radial null geodesic of the FRW metric obeys $dt/a=dr/\sqrt{1-K{r}^2}$. Using $1+z=1/a$ and $dz/dt=-(1+z)H(z)$ gives the radial comoving distanceFor $K>0$ this last integral is $\arcsin (\sqrt{K}{r}_1)/\sqrt{K}$, continued along the radial coordinate when necessary; for $K<0$ it is $\mathrm{arsinh}(\sqrt{|K|}{r}_1)/\sqrt{|K|}$. Inverting, with ${\Omega }_K=-K/H_0^2$, provesHere $S_K$ is sine for positive spatial curvature of an FLRW universe, hyperbolic sine for negative spatial curvature of an FLRW universe, and its argument for zero spatial curvature of an FLRW universe. The flat case is the continuous limit $d_L=(1+z)\chi$, not a literal division by zero. The original PDF has $H_0\sqrt{|{\Omega }_K|}$ in the denominator and $H=\dot{a}/a$; the converted TeX loses these details.

There is a genuine flux typo in the original PDF. A received bolometric radiative flux must have inverse-square luminosity distance, so the physically consistent thermal bremsstrahlung expression is $S_X\propto d_L^{-2}\int n_e^2{T}_e^{1/2}dV$. The printed single power of $d_L$ cannot imply the requested Hubble constant scaling.

For an approximately uniform, spherical galaxy cluster let its depth be $\ell =q\theta d_A$, where $q$ is a fixed geometrical factor. With fixed profile coefficients $A,B$, the Rayleigh-Jeans law limit of the thermal Sunyaev-Zeldovich effect and the bolometric X-ray radiative flux giveEliminate the electron number density and use $d_L=(1+z{)}^2{d}_A$:The cluster distance from Sunyaev-Zeldovich and X-ray signals is consequentlyAt low cosmological redshift, $d_A\simeq cz/H_0$, sowhen the other observables and profile parameters are held fixed. With the PDF's literal $d_L^{-1}$ instead, $S_X/D^2\propto \ell /(T_e^{3/2}{d}_L)=q\theta /[T_e^{3/2}(1+z{)}^2]$, independent of distance: this explicitly shows why the correction is necessary.

Real galaxy clusters are neither uniform nor generally spherical. The inferred depth depends on the ratio of line of sight depth to transverse extent; elongation and orientation selection therefore bias the distance. density clumps enhance X-ray emission, which weights $n_e^2$, more than the thermal Sunyaev-Zeldovich effect, which weights $n_e$. An unmodelled gas clumping factor greater than one makes the inferred distance too small and the inferred Hubble constant too large in this simple model. Temperature gradients and multiple thermal components bias the X-ray spectral temperature relative to the pressure-weighted Compton y parameter. Metals alter the cooling spectrum, and finite X-ray bands require a spectral and redshift correction to the bolometric approximation. Relativistic thermal corrections, the kinetic Sunyaev-Zeldovich effect, radio sources and primary Cosmic microwave background anisotropy contaminate the signal. Finally, calibration, angular-profile fitting and low-redshift peculiar velocities affect the observables and the relation between measured redshift and expansion distance.

The first equation is vertical hydrostatic equilibrium: the upward pressure gradient balances the central mass's vertical gravitational force. For $|z|\ll r$, the point-mass acceleration is $-GMz/(r^2+z^2{)}^{3/2}\simeq -{\Omega }_K^{2}z$, where Keplerian rotation gives ${\Omega }_K^2=GM/r^3$. The disc's self-gravity and vertical inertia are neglected.

The second equation is local thermal balance. The divergence of the vertical radiative flux equals viscous dissipation. In the convention of this alpha disk, $\nu =\alpha p/(\rho \Omega )$, and the dominant Keplerian shear has $rd\Omega /dr=-3\Omega /2$. ThereforeThe radiative flux increases from zero at the symmetric midplane to a positive outgoing value at the upper surface. Radial heat transport, thermal advection and rapid storage of thermal energy are excluded from this local balance.

The third equation is radiative diffusion, equivalentlyIt applies to optically thick material, with $\kappa$ the opacity per unit mass and $\sigma$ the Stefan-Boltzmann constant. Here constant electron-scattering opacity represents Thomson scattering; the formula uses a diffusive, nearly thermal radiation field. Positive upward flux requires temperature to decrease upwards.

The final equation is the ideal gas pressure law, with mean molecular weight ${\mu }_m$ in proton-mass units. Writing $\mathcal{R}=k/({\mu }_m{m}_p)$ gives $p=\mathcal{R}\rho T$. The pressure is gas pressure because radiation pressure is assumed negligible. The four equations express vertical support, local heating balance, diffusive cooling and the gas equation of state.

In a spatially flat FLRW metric, a radial null geodesic satisfies $d\chi /dt=\pm c/a(t)$. For a source and observer at fixed comoving separation, $\chi ={\int }_{t_e}^{t_0}cdt/a(t)$. Applying this to successive wave crests gives $d{t}_0/a(t_0)=d{t}_e/a(t_e)$ and therefore the cosmological redshift $1+z=a(t_0)/a(t_e)$.

Each photon's energy is reduced by $1+z$, and the arrival interval is increased by $1+z$. The received power crossing the observation sphere is consequently $L/(1+z{)}^2$. Its physical area is $4\pi a(t_0{)}^2|x{|}^2$, so the radiative flux isThis is the luminosity distance relation with $d_L=(1+z)a(t_0)|x|$. The Euclidean area formula presumes spatial flatness; in a curved FLRW metric the transverse comoving distance replaces $|x|$.

For constant opacity, the ratio $T^3/\rho =3c\lambda \mathcal{R}/(4\sigma )$ cancels the height dependence in the radiative diffusion coefficient. Since $T_z=-{\Omega }^{2}z/[4(1+\lambda )\mathcal{R}]$, the radiative flux isConsequently $F_z=c\lambda {\Omega }^2/[\kappa (1+\lambda )]$. Equating this to the viscous heating rate givesThus the dynamic viscosity is uniform in height even though pressure and mass density decrease. Its units are mass per length per time, since $c/\kappa$ has those units. A pointwise pressure-proportional alpha prescription would be incompatible with this exact profile, which is why the subsequent alpha relation is vertically averaged.

The first relation is vertical hydrostatic equilibrium. In a thin disc, the central object's gravity near the midplane is $g_z=-{\Omega }^{2}z$; the pressure gradient balances it. The second is an equation of state combining ideal gas pressure with isotropic radiation pressure. Write $\mathcal{R}=k/({\mu }_m{m}_p)$, so $p_g=\mathcal{R}\rho T$ and $p_r=4\sigma T^4/(3c)$; $\sigma$ is the Stefan-Boltzmann constant and ${\mu }_m$ the mean molecular weight in proton-mass units.

The third is local energy balance: the increase of upward radiative flux with height equals viscous dissipation. Here $\mu$ is dynamic viscosity, not mean molecular weight. A Keplerian disc has shear $rd\Omega /dr=-3\Omega /2$, so its heating rate is $\mu (r{\Omega }^{\prime }{)}^2=9\mu {\Omega }^2/4$. The fourth is optically thick radiative diffusion, with opacity per unit mass $\kappa$ and diffusion directed down the temperature gradient. Midplane symmetry gives $F(0)=0$ and oppositely directed fluxes from the two faces. The four relations express vertical force balance, gas–radiation thermodynamics, local heating, and diffusive cooling.

Write $u=\cos \theta$, $S=j/\kappa$ and use axial symmetry. The radiative flux is $F=2\pi {\int }_{-1}^{1}uIdu$. Integrating the radiative transfer equation over directions givesThere are no energy sources in the atmosphere, so radiative equilibrium makes $F$ constant. Consequently

For the linear angular approximation $I=A+Cu$, the radiation-field moments areThus it satisfies the Eddington closure approximation, andSince $F$ is constant, $C^{\prime }=0$. Substitution in the radiative transfer equation, using $S=A$, gives $u{A}^{\prime }=Cu$, hence $\boxed{A^{\prime }=C}$.

In this moment approximation, the inward hemispheric flux at the surface isSetting this boundary moment to zero gives $A_0=2C/3$. ThereforeCombining it with the thermal source function and $F=\sigma T_e^4$ gives the grey atmosphere temperature law:

The surface condition is an approximate angular-moment condition. If the affine intensity were interpreted as an exact nonnegative boundary radiation field, no incident radiation would require $I(0,u)=0$ for every $u<0$; a linear polynomial vanishing on that interval would have $A_0=C=0$ and no outward flux either. The derived formal profile indeed has $I(0,-1)=-F/(4\pi )<0$. This positivity limitation of a linear Eddington intensity is why one uses the Eddington surface boundary condition as a moment closure, not the affine ansatz as an exact emergent angular intensity. An outward direction-independent intensity with no inward intensity gives the same boundary moments.

For the gas-pressure-dominated atmosphere, combine $dP/dz=-\rho g$ with $d\tau /dz=-\kappa \rho$. Thus the hydrostatic equilibrium in optical depth is $dP/d\tau =g/\kappa$. Let $w=T^4$. The grey atmosphere law implies $dw/d\tau =3T_0^4/2$. With the specified power-law opacity,At the outer vacuum limit take $P=0$ and $T=T_0$. Since $\alpha ,\beta >0$, integration givesThis is the grey-atmosphere pressure with power-law opacity relation.

At the matching radius the local radiative flux is $L_r/(4\pi r^2)=\sigma T^4$. In the thin plane-parallel atmosphere approximation this equals $F=\sigma T_e^4$, so $T=T_e$ and $T^4=2T_0^4$. Multiplying the pressure relation by ${\kappa }_0{T}^{4-4\beta }/g$ yields the required surface pressure-opacity relationHere $P\kappa$ is a product, not a pressure subscript.

For unresolved, uniformly bright blackbodies at a common distance $d$, the monochromatic observed radiative fluxes are $F_{s,\lambda }=\pi B_{\lambda }(T_s)R_s^2/d^2$ and $F_{p,\lambda }=\pi B_{\lambda }(T_p)R_p^2/d^2$. Define the planet-star contrast $r_{\lambda }=F_{p,\lambda }/F_{s,\lambda }$. The Planck law givesOutside the exoplanet secondary eclipse, the system flux is $F_s+F_p$; during complete occultation it is $F_s$. Consequently the exact thermal eclipse depth normalized to the out-of-eclipse light isOnly in the faint-planet limit is $d_{\lambda }\simeq r_{\lambda }$. Reflected starlight, partial occultation and spatial temperature variations are omitted under the stated blackbody model.

For a cooler hot Jupiter with $T_p<T_s$, the planet-star contrast is exponentially small at short wavelengths and increases towards a constant plateau. This is the monotonicity of blackbody planet-star contrast, rather than the peaked shape of the planet's absolute blackbody flux. An illustrative example uses $T_p=1500\mathrm{K}$, $T_s=5800\mathrm{K}$ and $R_p/R_s=0.1$:

At wavelengths long enough for the Rayleigh-Jeans law to apply to both bodies, $B_{\lambda }\simeq 2c{k}_{B}T/{\lambda }^4$. ThereforeThe common ${\lambda }^{-4}$ dependence cancels. In the illustrative case, the contrast plateau is about $2.6\times 10^{-3}$, or $2600$ parts per million.

Let $z$ increase outward and define the inward optical depth by $d\tau /dz=-\bar{\kappa }\rho$. With constant outward internal radiative flux $F=\sigma T_{\mathrm{int}}^4$, radiative diffusion givesThe constant $q$ is a boundary condition; diffusion alone does not determine it. For an unirradiated grey atmosphere with the Eddington closure approximation and Eddington surface boundary condition, $q=2/3$, soThe internal effective temperature of a planet is defined by its intrinsic cooling flux, not by its incident stellar heating. Since $dT/d\tau >0$ and $d\tau /dz<0$, temperature decreases outward. Towards the thin upper layers, $T\to 2^{-1/4}{T}_{\mathrm{int}}\simeq 0.84T_{\mathrm{int}}$; if the density and optical-depth gradient vanish there, $dT/dz\to 0$. This is the upper nearly isothermal atmosphere. The diffusion approximation itself fails at small optical depth: the grey boundary closure supplies the approximate continuation. Small irradiation perturbs this intrinsic-flux solution.

Young, self-luminous gas giants at wide orbital separations can have intrinsic cooling dominate their photospheric budget. They are favorable for exoplanet direct imaging, especially in the infrared, where their own thermal radiation is easier to separate from the host light. A strongly irradiated hot Jupiter instead has a large stable outer radiative region set mainly by stellar heating; it can be nearly isothermal over a broad pressure range or develop an atmospheric thermal inversion if stellar light is absorbed sufficiently high. Inversion is not inevitable for every strongly irradiated planet. Deep convection begins below its radiative-convective boundary.

The intrinsic curve follows the grey formula. The irradiated curves illustrate possible shapes only; they are not solutions for a specified opacity model. Smaller optical depth corresponds to greater altitude.

Assume a circular orbit of radius $a$, a spherical planet, bolometric Bond albedo $A_B$, negligible internal heat, unit thermal emissivity and complete redistribution of absorbed heat over the sphere. The stellar luminosity is $L_s=4\pi R_s^2\sigma T_s^4$, and the planet absorbs the incident radiative flux through its projected cross-section:Thermal reradiation is $P_{\mathrm{emit}}=4\pi R_p^2\sigma T_{\mathrm{eq}}^4$. Equating these gives the planetary equilibrium temperatureThe planet's radius cancels. A non-black thermal emissivity $ϵ$ divides the absorbed flux by $ϵ$ in the fourth-power balance. For uniform dayside-only reradiation, the emitting area is $2\pi R_p^2$ and $T_{\mathrm{day}}=2^{1/4}{T}_{\mathrm{eq}}$. With no local redistribution, the substellar point has $T_{\mathrm{sub}}=\sqrt{2}{T}_{\mathrm{eq}}$, while other points depend on incidence angle. These are different temperature conventions, not contradictory formulas.

The planetary equilibrium temperature is not the surface greenhouse temperature or the internal effective temperature of a planet. If intrinsic cooling matters and both powers escape through the same emitting area, the total effective temperature obeys $T_{\mathrm{eff}}^4=T_{\mathrm{int}}^4+T_{\mathrm{eq}}^4$.

The four equations describe, respectively, vertical hydrostatic equilibrium in the central object's gravity, the vertical divergence of radiative flux balancing viscous dissipation, optically thick radiative diffusion, and the equation of state combining ideal gas gas pressure with radiation pressure. Here $\mu$ is dynamic viscosity, not the mean molecular weight ${\mu }_m$; $\kappa$ is the opacity, $\sigma$ the Stefan-Boltzmann constant, and $\Omega$ the local orbital frequency.

In the radiation-pressure limit, $p=4\sigma T^4/(3c)$, so radiative diffusion givesCompare this with vertical hydrostatic equilibrium. Wherever $\rho \ne 0$,At fixed radius, constant opacity makes $dF/dz=c{\Omega }^2/\kappa$. The viscous dissipation equation therefore yieldsindependent of height within the optically thick radiation-supported layer. The result assumes the supplied gravity, heating and diffusion equations; optically thin surface corrections are outside that approximation.

A natural interpretation of the short-wavelength peak is reflected starlight, while the longer-wavelength peak is planetary thermal radiation. A reflected spectrum approximately follows the stellar spectrum multiplied by the wavelength-dependent geometric albedo; thermal radiative flux approximately follows the planet's Planck function, modulated by molecular opacity. Thus two peaks need not represent two planetary surface temperatures.

For a quantitative estimate assume both are broad $F_{\lambda }$ peaks, reflection has a slowly varying geometric albedo, and star and planet have approximately blackbody spectral envelopes. Wien's displacement law then givesThe stellar estimate is compatible at order of magnitude with an old, relatively small main-sequence star; equal age does not mean equal temperature to the Sun. Assume the far-infrared signal is thermal and sufficiently long-wavelength that the Rayleigh-Jeans law applies to both bodies. The planet-star radius estimate in the Rayleigh-Jeans limit givesand thereforeThis is a small volatile-rich-planet-sized estimate, not a Jupiter-sized one. It is conditional: peaks caused by molecular windows, strongly chromatic reflection, or a spectrum expressed as $F_{\nu }$ rather than $F_{\lambda }$ do not support those two Wien estimates. Without $T_p/T_{\ast }$, the far-infrared ratio only fixes $R_p^2{T}_p/T_{\ast }$.

For a close-in hot planet around a small star, transit-based atmospheric observations are the natural route if the orbital geometry allows them. Exoplanet transmission spectra gain from the small stellar radius, with a limb signal scaling as $2R_{p}H/R_{\ast }^2$; they probe composition at the day-night terminator. The stated $10^{-3}$ thermal contrast also makes exoplanet secondary eclipse measurements a particularly useful route to the dayside exoplanet emission spectrum. Resolving such a close-in small planet by exoplanet direct imaging is much harder. Transit and secondary-eclipse spectroscopy are favoured for a transiting close-in interpretation; the supplied spectrum alone does not determine the orbital geometry or a unique best technique.

Assume the present epoch has the same age as present-day Jupiter, the host now has solar luminosity, and moving inward did not change the stipulated intrinsic cooling law or its normalization. This neglects persistent tidal heating and irradiation-induced suppression of cooling. The intrinsic planetary luminosity is then approximately $L_{\mathrm{int},\mathrm{J}}$, not its value at the migration epoch.

Define incident irradiation as intercepted power before reflection,Comparing with present-day Jupiter at $a_J\simeq 5\mathrm{AU}$ givesFor equal radii,Retaining unequal radii multiplies the right side by $(R_J/R_p{)}^2$. If irradiation instead denotes incident radiative flux per area, the orbital factor is still $10^4$, but it must be compared with an intrinsic flux consistently. Absorbed power also includes $1-A_B$, so comparisons of absorbed irradiation require the two Bond albedos.

Close-in gas giants motivate planetary migration when compared with formation models. Two useful observational diagnostics are then orbital eccentricities and spin-orbit geometry. An eccentric population of wider potential progenitors together with circular short-period orbits supports eccentricity excitation followed by tidal dissipation. The second diagnostic is stellar obliquity, including misaligned or retrograde orbits measured through the Rossiter-McLaughlin effect; these can favour scattering or secular pathways over smooth coplanar migration. Conversely, aligned resonant architectures are compatible with disc-driven migration. None is unique: primordial disc tilt or alternative formation can mimic some signatures. The expected present-day ratio is suppressed by the inverse-square orbital factor, while eccentricities and spin-orbit geometry test migration pathways.

Take an unresolved spherical planet at Euclidean distance $d\gg R$, with uniform surface temperature $T$, stationary isotropic emission, and no intervening extinction. Isothermality does not by itself imply blackbody radiation; first derive the geometric result for a uniform emergent specific intensity $I_{\lambda }$.

The observer sees the projected disc with apparent solid angle ${\Omega }_p\simeq \pi R^2/d^2$. Integrating its specific intensity over that angle gives the observed spectral radiative fluxwhere ${\theta }_{\mathrm{obs}}$ is the small angle to the detector normal. For a blackbody, $I_{\lambda }=B_{\lambda }(T)$, and the Stefan–Boltzmann law gives ${\int }_0^{\infty }\pi B_{\lambda }(T)d\lambda ={\sigma }_{\mathrm{SB}}{T}^4$. HenceEquivalently, the surface radiative flux is ${\sigma }_{\mathrm{SB}}{T}^4$ and the luminosity of a spherical blackbody is $L=4\pi R^2{\sigma }_{\mathrm{SB}}{T}^4$, leading to the same result by inverse-square dilution. A uniform wavelength-dependent emissivity ${ϵ}_{\lambda }$ multiplies the spectral formula; an atmospheric spectrum can require a full angular radiative transfer integral rather than a single Planck function.