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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 $P\kappa /g=4\alpha \log 2/3$.

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.

Use the notation $I_{\nu }(\mu ,\tau )$ throughout, with optical depth increasing inward and $0<\mu \le 1$ for an outward ray. This removes the change in argument order used in the printed surface-intensity notation. Multiplying the radiative transfer equation by its integrating factor givesIntegrating from the surface to a deeper boundary $b$ yieldsFor a semi-infinite atmosphere with no exponentially growing homogeneous term, the boundary term vanishes as $b\to \infty$. The formal solution of the radiative transfer equation is thereforeThe usual deep-atmosphere boundary condition is necessary; a first-order differential equation alone does not fix the emerging specific intensity.

For a linear source function, the zeroth and first exponential moments give $I_{\nu }(\mu ,0)=a_0+a_1\mu$. Equivalently $I_{\nu }(\mu ,0)=S_{\nu }(\tau =\mu )$, the Eddington-Barbier relation. If the source rises inward, rays near the limb sample cooler, shallower layers, producing limb darkening.

For the finite polynomial, put $u=t/\mu$ and integrate each term. Repeated integration by parts gives ${\int }_0^{\infty }{u}^j{e}^{-u}du=j!$, so the polynomial source function moments for emergent intensity giveThe limiting grazing-ray specific intensity is $a_0$. If the displayed polynomial is only a local Taylor approximation to a more general source, this formula applies to that approximation and a source-function remainder also contributes; a finite local expansion is not automatically an exact description at every optical depth.

For a constant source in an overlying layer of optical thickness $\tau$, the radial ray has the exact transfer relationExpanding the exponentials gives the constant source transfer through a thin layer:This finite-layer result does not require the specific intensity below the layer to equal its source function. The sign of $S_{\nu }-I_{\nu }(1,\tau )$ already distinguishes emission from absorption.

An absorption line forms when a bound-bound transition increases opacity at selected frequencies. In a photosphere whose temperature decreases outward, optical depth of order one at the line frequency occurs higher than at a nearby continuum frequency. In an absorption-dominated region in local thermodynamic equilibrium, the source is the Planck function; the cooler line-forming layer then emits less specific intensity than the deeper continuum-forming layer. The result is a dark spectral feature. Scattering and departures from LTE modify the source, so this temperature-gradient picture is a useful mechanism rather than a universal identity for every line.

Two distinct situations can produce emission lines. First, a heated chromosphere has an outward temperature rise, so an optically thick line can sample gas hotter than the continuum photosphere beneath it. Secondly, a hot optically thin circumstellar envelope or stellar wind emits line photons through collisional excitation or recombination, with no comparably bright background along all rays; its added line flux can exceed the underlying continuum. Winds of hot luminous stars can also give a mixture of emission and absorption across a Doppler-broadened line profile. These examples explain the change in specific intensity through either an enhanced source function or additional emitting material.

Take the optical depth to increase inward and let $\mu >0$ denote an outward ray. For thermal absorption and emission in local thermodynamic equilibrium, the radiative transfer source function is the Planck function, so the radiative transfer equation isThis assumes negligible scattering, or a source function genuinely thermalized to $B_{\lambda }$; LTE by itself does not turn an arbitrary scattering source into a Planck function. Multiply by $e^{-\tau /\mu }$, integrate between the two boundaries, and solve for the outward specific intensity:The bottom boundary value is the quantity denoted $I_{\lambda }(0)$ in the supplied notation; it is not an optical-depth-zero boundary. The first term is attenuated incident radiation, and the second is emission from the intervening layers. This is the formal solution of the radiative transfer equation.

For an isothermal atmosphere at temperature $T$, the Planck function is constant and the integral can be evaluated:For a bounded bottom intensity and ${\tau }_1\to \infty$, $I_{\lambda }=B_{\lambda }(T)$ in every outgoing direction, and $F_{\lambda }=\pi B_{\lambda }(T)$. Thus the emergent spectrum is a blackbody spectrum. The blackbody limit of an isothermal atmosphere does not require a temperature gradient: it follows from complete thermalization in an optically thick medium.

Take $R_p$ to be the radius of an opaque planetary disc and $H$ the geometric thickness of the model atmosphere, not necessarily one atmospheric scale height. Assume a uniform stellar specific intensity, a fully projected non-grazing transit, negligible planetary emission in the measured band, and no scattering or refraction returning light to the beam. The cylindrical approximation assigns the same slant optical depth ${\tau }_{\lambda }$ to every ray through the annulus.

The opaque disc blocks area $\pi R_p^2$. The annulus has projected area $\pi [(R_p+H{)}^2-R_p^2]$, and the radiative transfer equation transmits fraction $e^{-{\tau }_{\lambda }}$ through it. Its blocked fraction is therefore $1-e^{-{\tau }_{\lambda }}$. Dividing the missing light by the unobscured stellar-disc light gives the exoplanet transmission spectrumFor $H\ll R_p$, the annulus model for transmission spectroscopy becomesThe optically thin excess is $2R_{p}H{\tau }_{\lambda }/R_s^2$; the optically thick limit is the area ratio $(R_p+H{)}^2/R_s^2$ before the thin-annulus approximation. For a realistic atmosphere, the slant optical depth varies with the ray's impact parameter $b$, giving insteadLimb darkening replaces the simple area weighting by the local stellar specific intensity. The constant-depth model is therefore an explicit geometric approximation, not the slant-depth law of a spherical hydrostatic atmosphere.

Consider a narrow bundle of rays between two surface elements $d{A}_1,d{A}_2$ separated by $s$ in a transparent, stationary Euclidean vacuum. Let the ray make angles ${\theta }_1,{\theta }_2$ to the respective surface normals. The receiving element subtends solid angle $d{\Omega }_1=\cos {\theta }_{2}d{A}_2/s^2$ at the emitter. By the definition of specific intensity, the beam power in frequency interval $d\nu$ isAt the receiving end the source subtends $d{\Omega }_2=\cos {\theta }_{1}d{A}_1/s^2$, so the identical power isIn the absence of absorption, emission or frequency shifts, conservation of energy equates these expressions:This vacuum conservation of specific intensity also follows directly from the radiative transfer equation $d{I}_{\nu }/ds=0$. The apparent solid angle of an unresolved source shrinks as $s^{-2}$, so its integrated observed radiative flux still follows the inverse-square law. Specific intensity and unresolved flux are different quantities.

The vacuum and frequency assumptions are essential: absorption and emission change specific intensity, and gravitational or cosmological redshift conserves $I_{\nu }/{\nu }^3$ along a ray instead of $I_{\nu }$ itself. A fixed telescope aperture observing an unresolved object also cannot infer constant total received power from intensity conservation.

At the stellar surface boundary condition, match the interior to a stellar atmosphere, rather than setting the temperature and pressure to zero at an arbitrary radius. At a photosphere of radius $R$, the enclosed mass is $M$, the luminosity is $L$, and the outward radiative flux defines the effective temperature through $F=L/(4\pi R^2)={\sigma }_{\mathrm{SB}}{T}_{\mathrm{eff}}^4$. With no external illumination the incoming specific intensity satisfies $I_{\nu }(0,\mu )=0$ for $\mu <0$. A deeper boundary is matched to the nearly isotropic radiative diffusion field. A grey atmosphere commonly matches near $\tau =2/3$; under the Eddington surface boundary condition this gives $T=T_{\mathrm{eff}}$. For roughly constant gravity and opacity, the total pressure rises by $g\tau /\kappa$ relative to its outer boundary value; if radiation supplies appreciable support, the gas-pressure gradient uses the effective gravity $g-\kappa F/c$ instead.

Take $z$ increasing outward and $\mu$ the outward direction cosine. If $j_{\nu }$ is emission per unit path length and solid angle, the radiative transfer equation isDefine inward-increasing optical depth by $d{\tau }_{\nu }=-\rho {\kappa }_{\nu }dz$ and the source function by $S_{\nu }=j_{\nu }/(\rho {\kappa }_{\nu })$. Then $\mu {\partial }_{{\tau }_{\nu }}{I}_{\nu }=I_{\nu }-S_{\nu }$. This states the emission-coefficient convention; if $j_{\nu }$ is defined per unit mass instead, the emission term is $\rho j_{\nu }$.

For frequency-integrated intensity, introduce the radiation-field momentsThe first angular approximation writes $I\simeq J+3\mu H$. Its odd part does not contribute to $K$, soThis is the Eddington closure approximation, not an exact description of the escaping angular distribution. An approximate surface condition takes the outgoing hemisphere to have direction-independent intensity $I_0$ and the incoming hemisphere to be dark. Its moments give $F=\pi I_0$ and $J(0)=I_0/2=F/(2\pi )$. Integrating the moment equations with the closure then gives $J=3F(\tau +2/3)/(4\pi )$; in thermal equilibrium, $J=a_{\mathrm{rad}}c{T}^4/(4\pi )$ gives $T^4=(3/4)T_{\mathrm{eff}}^4(\tau +2/3)$. This derives the usual approximate photospheric matching condition rather than imposing a zero surface temperature. For conservative coherent isotropic scattering, $S=J$. Integrating the transfer equation gives $dH/d\tau =0$ and $dK/d\tau =H$, the moment conditions for radiative equilibrium.

The exponential ansatz in the PDF has a genuine consistency problem. Compute its moments before making any approximation. For a nonsingular real intensity on all $-1\le \mu \le 1$, take $|ϵ|<1$ and defineThe continuous zero limits are $A_0=1$, $A_1=0$, $A_2=1/3$. Direct integration yieldsFor a decaying nontrivial correction, $\delta >0$ and $\gamma \ne 0$. Exact constant flux then requires $A_1=0$. But $\mathrm{atanh}ϵ/ϵ>1$ for every nonzero real $|ϵ|<1$, so $A_1=0$ forces $ϵ=0$. Thus exact radiative equilibrium cannot imply a small nonzero $ϵ$ for this ansatz.

Substitution into the full transfer equation makes the obstruction sharper. The exponential coefficient must satisfy, for every $\mu$,Comparing the constant and linear coefficients gives $A_0=1$ and $\delta =A_0ϵ=ϵ$. Hence regular real coefficients require $\delta =ϵ=0$, eliminating any decaying correction. This is the obstruction to a single smooth exponential mode in conservative grey transfer. There is no exact nontrivial real solution of the stipulated form with a small nonzero decay parameter. Choosing, for example, $ϵ=\delta =0.1$ and $\gamma \ne 0$ supplies an explicit counterexample to the claimed exact flux constancy, because $A_1\ne 0$.

Several requested conclusions remain useful as formal approximations, provided they are identified as such. If $\delta >0$, the large-depth exponential vanishes, giving $F\to 4\pi \alpha /3$, hence $\alpha =3F/(4\pi )$ for the asymptotic physical flux. It also gives $c{P}_{\mathrm{rad}}\to (4\pi /3)J$. Since $A_0=1+{ϵ}^2/3+O({ϵ}^4)$, neglecting second-order angular errors yields the formal relation $\delta \simeq ϵ$. However $A_1=-ϵ/3+O({ϵ}^3)$, so the finite-depth flux still varies at first order. The formal relation is not an exact transfer or constant-flux solution.

There is also a flux-normalization mismatch in the PDF's last displayed relation. With the same physical $F$ used above, the Hopf function of a grey atmosphere is defined byThe PDF's coefficient $3F/4$ would require redefining its $F$ as physical flux divided by $\pi$; it cannot simultaneously use the earlier $\alpha =3F/(4\pi )$ for the same physical flux. With the consistent physical-flux normalization, formal moment matching givesIn the intended small-$ϵ$ approximation this isThe approximate outgoing surface ansatz then gives the limb darkening ratioFor finite $ϵ$, $\gamma =-0.14/A_0$ gives a parameter-dependent ratio, not a unique exact value. The two Hopf endpoint values alone do not determine the exact emergent intensity: the formal solution of the radiative transfer equation uses the entire source profile, $I(0,\mu )={\int }_0^{\infty }J(t)e^{-t/\mu }dt/\mu$ for $\mu >0$. Nor does the simple linear-plus-exponential ansatz satisfy the exact no-incoming-radiation condition for every negative $\mu$. Thus $2.75$ is a clearly labelled approximate disk-centre-to-limb result, not an exact consequence of inconsistent premises.

The outward radiative flux is $F=L/(4\pi r^2)$. A photon of energy $E$ carries momentum $E/c$, so the momentum transferred per unit volume and time to matter is $\kappa \rho F/c$, when $\kappa$ is the appropriate transport opacity. Momentum conservation makes the radiation stress decrease outward by this amount. In the thin atmosphere, where the radiation stress is treated by its scalar radiation pressure,This is the first moment of the radiative transfer equation. The grey atmosphere approximation uses a frequency-independent transport opacity; in an optically thick diffusion regime it becomes the Rosseland mean opacity. Near the surface the plane-parallel atmosphere approximation removes geometric stress-divergence corrections.

Assume a uniform stellar disc of radius $R_{\star }$, an opaque planetary radius $R_p$, and a thin atmospheric annulus of thickness $z\ll R_p$. Neglect planetary light during transit and scattering back into the beam. The attenuation part of the radiative transfer equation gives $I_{\lambda }=I_{\lambda ,0}{e}^{-{\tau }_{\lambda }}$, where ${\tau }_{\lambda }$ is the slant optical depth along a stellar ray through the atmosphere. The fraction of light removed from that annulus is therefore $1-e^{-{\tau }_{\lambda }}$.

If the annulus is represented by one effective optical depth, its area divided by the stellar area isThe annulus model for transmission spectroscopy consequently gives the extra transit depthHere ${\delta }_{\lambda }$ is normalized to the unobscured stellar flux and excludes the opaque-disc depth $(R_p/R_{\star }{)}^2$. For a spectral feature measured relative to a continuum with slant optical depth ${\tau }_c$, the corresponding contrast is $A(e^{-{\tau }_c}-e^{-{\tau }_{\lambda }})$; the displayed formula takes the annular continuum to be transparent.

A real exoplanet transmission spectrum has an impact-parameter-dependent optical depth. Its more accurate expression isStellar limb darkening and horizontally varying clouds further modify the weighting. The single-annulus approximation is useful for estimating an atmospheric spectral-feature amplitude, rather than predicting every spectral line.

Assume local thermodynamic equilibrium, negligible scattering of thermal radiation, and a plane-parallel atmosphere. With inward optical depth ${\tau }_{\lambda }$ and outward ray cosine $\mu$, the radiative transfer equation isFor a deep atmosphere, its formal solution of the radiative transfer equation isThe weighting samples ${\tau }_{\lambda }$ of order unity. The Eddington-Barbier relation makes this precise when the source function is nearly linear: $I_{\lambda }(0,\mu )\simeq B_{\lambda }[T({\tau }_{\lambda }=\mu )]$.

A molecular band with greater opacity reaches unit optical depth higher than the nearby continuum. If temperature decreases upward, that band samples cooler gas and appears in absorption. If an atmospheric thermal inversion makes the upper gas hotter, the band appears in emission. If both depths have the same temperature, an opaque isothermal atmosphere hasand no opacity-dependent features. This last conclusion assumes all wavelengths are opaque or the lower boundary emits the same Planck function; an optically thin isothermal layer over a different background need not be featureless.

For a plane-parallel atmosphere with optical depth increasing downward, the radiative transfer equation isIntegrating over solid angle gives the first radiation-field momentIn radiative equilibrium, matter has no net local radiative heating, so the opacity-weighted frequency integral of $J_{\nu }-S_{\nu }$ vanishes. After converting each optical-depth derivative to physical depth and integrating over frequency,Thus the bolometric internal flux is constant with depth, as stated by constant flux in a plane-parallel radiative-equilibrium atmosphere.

In local thermal equilibrium $S_{\nu }=B_{\nu }(T)$. Define the planet's internal effective temperature of a planet by the constant outward flux. The standard Planck law integral givesThe corresponding intrinsic luminosity is $L_{\mathrm{int}}=4\pi R_p^2\sigma T_{\mathrm{eff}}^4$.

Interstellar dust both hides and reveals galaxies. Absorption and scattering remove short-wavelength photons from a line of sight, causing interstellar extinction, reddening, anisotropic scattering, and polarization. These effects obscure embedded star formation, bias luminosities, colours, stellar masses, and star-formation rates, and can make an edge-on or dusty galaxy appear older and fainter. Dust also converts absorbed ultraviolet and optical power into far-infrared and submillimetre thermal radiation. That reradiation exposes otherwise hidden star formation, constrains dust temperature and mass, and helps map cold molecular material; polarization traces magnetic-field orientation. Grain surfaces also catalyse molecular-hydrogen formation, while photoelectric emission from grains heats interstellar gas.

One spherical grain has emitting area $4\pi a^2$. Since isotropic blackbody intensity $B_{\nu }$ gives surface flux $\pi B_{\nu }$, its spectral luminosity isFor $N=M_d/m_d$ optically thin grains at distance $D$, the inverse-square law givesDefining the grain mass absorption coefficientthe optically thin dust-mass estimator is

For a constant source function and no incident background, the radiative transfer equation integrates toIf the cloud subtends solid angle $\Omega$, uniform intensity givesThe physical projected area is $A=\Omega D^2$. Its grain column density is $N/A=M_d/(m_d\Omega D^2)$, so the dust optical depth isEliminating ${\tau }_{\nu }$ gives the finite-optical-depth resultIt requires $F_{\nu }<\Omega B_{\nu }$, as demanded by the blackbody brightness limit. When ${\tau }_{\nu }\ll 1$, $1-e^{-{\tau }_{\nu }}={\tau }_{\nu }+O({\tau }_{\nu }^2)$, so $F_{\nu }\simeq \Omega B_{\nu }{\tau }_{\nu }={\kappa }_{\nu }{M}_d{B}_{\nu }/D^2$ and the result reduces to the optically thin estimator.

For the plane-parallel radiative transfer equationangular integration gives the zeroth radiation-field moment equationThe net radiative heating per unit volume is thereforeRadiative equilibrium requires it to vanish, equivalently that the frequency-integrated radiative flux be independent of depth:In local thermodynamic equilibrium with coherent isotropic scattering,where ${\omega }_{\nu }$ is the single-scattering albedo. Since ${\alpha }_{\nu }(1-{\omega }_{\nu })={\alpha }_{\nu ,\mathrm{abs}}$, the condition becomesConservative scattering redistributes directions but contributes no net material heating.

Along a ray, the radiative transfer equation has the formal solution of the radiative transfer equationfor each isothermal region in local thermodynamic equilibrium, neglecting scattering. For a self-luminous atmosphere with no incident intensity from below at the relevant photosphere, and in the optically thick limit, $I_{\nu ,i}=B_{\nu }(T_i)$.

The observed radiative flux is the projected-disc integralThe inner region occupies projected fraction $f={\sin }^2{\theta }_T$, henceFor finite optical depths, each Planck law function in this formula is replaced by its corresponding emergent intensity above.

Put $\mu =\cos \theta$ and $S=j/\kappa$. Integrating the radiative transfer equation over solid angle givesThe first integral is the radiative flux $F$. A stationary atmosphere with no local energy source has $dF/d\tau =0$, soBy definition, the mean intensity is $J=(4\pi {)}^{-1}\int Id\Omega$. Thereforewhich is radiative equilibrium: the angle-integrated emission and absorption rates are equal.

Without scattering and in local thermodynamic equilibrium, the radiative transfer equation has the emergent solutionThe Eddington-Barbier relation gives the useful approximationA molecular band has larger opacity than its neighboring continuum and therefore reaches optical depth unity at lower pressure.

If temperature decreases outward, the band samples cooler gas and appears in absorption. If the atmosphere is isothermal, both levels have the same source function and the feature disappears. If an atmospheric thermal inversion makes the upper layer hotter, the band appears in emission. For a weak separation of formation pressures,which explicitly shows that feature sign and amplitude measure the vertical temperature gradient.

In a transparent, stationary vacuum with Euclidean geometry and observers at rest in the same frame, the radiative transfer equation has zero absorption and emission, so $d{I}_{\nu }/ds=0$ along a light ray. The same specific intensity reaches the observer without geometric dilution. More generally, $I_{\nu }/{\nu }^3$ is a Lorentz invariant; collisionless propagation preserves this ratio along a ray even with gravitational or cosmological redshift, rather than preserving $I_{\nu }$ at shifted frequencies. ^[ref].

The elementary geometry is instructive. Two facing surface elements separated by distance $s$ exchange spectral radiant fluxThe receiver subtends $d{\Omega }_2=\cos {\theta }_{2}d{A}_2/s^2$ at the source. Describing that same traveling beam at the receiver instead uses $d{\Omega }_1=\cos {\theta }_{1}d{A}_1/s^2$. Both descriptions have identical geometric factors, so conservation of its radiant flux gives equal specific intensities at the two ends. This compares the same propagation direction at both ends, not emission in the opposite direction.

A uniform source remains equally bright per resolved direction under these assumptions, but occupies a smaller apparent solid angle farther away. For a small source of projected area $A_{\perp }$ at distance $d$ and a small receiver facing it, irradiance is approximately $E=L{\Omega }_{\mathrm{source}}\simeq L{A}_{\perp }/d^2$. The radiant flux intercepted by receiver area $A_d$ is approximately $E{A}_d$. With $L=10\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$, $A_{\perp }=0.020{\mathrm{m}}^2$ and $A_d=10^{-4}{\mathrm{m}}^2$, distances $10$ and $20\mathrm{m}$ give $E\simeq 0.002$ and $0.0005\mathrm{W}{\mathrm{m}}^{-2}$, and received radiant fluxes of $0.20$ and $0.050\mu \mathrm{W}$. The inverse-square law concerns collected radiation from the whole source; $L$ remains $10$ at either distance.