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The bulk-parameter table gives the Moon's volumetric mean radius as 1737.4 km and solar irradiance as $1361\mathrm{W}{\mathrm{m}}^{-2}$. It separately lists Bond albedo 0.11 and geometric albedo 0.12; these are different measures of reflection.

For a small, uniform surface at temperature $T$, enclosed by much larger black walls all at temperature $T_s$, the outward net heat transfer by thermal radiation isAssume an opaque grey body that is a diffuse thermal emitter, with the same emissivity $ϵ$ and radiometric absorptance at the surface's current temperature. Its emitting area $A$ sees only the enclosure: the view factor to the walls is one, with no self-viewing surface, opening or intervening obstruction. The intervening medium must neither absorb nor emit appreciable thermal radiation; a vacuum supplies a useful idealization. The enclosure is maintained at uniform $T_s$. The small-body assumption makes that reservoir description physically reasonable; genuinely black isothermal walls supply the incoming irradiance even without an infinite area ratio. Positive ${\dot{Q}}_{\mathrm{rad}}$ means net loss by the surface; negative means net gain. This expression counts thermal radiation alone, not all heat transfer.^[ref]

The reasoning is a two-way energy balance. Black walls supply irradiance $E=\sigma T_s^4$, so the surface absorbs $P_{\mathrm{abs}}=ϵAE$ and emits $P_{\mathrm{emit}}=ϵA\sigma T^4$. Subtract these to obtain ${\dot{Q}}_{\mathrm{rad}}$. For an opaque surface the remaining fraction $1-ϵ$ of incoming radiant flux is reflected; counting emitted plus reflected outgoing radiant flux minus total incoming radiant flux gives the same result. Kirchhoff's law of thermal radiation justifies equal spectral absorption and emission factors; the grey body assumption lets one factor be taken outside both spectral integrals.

Choose $ϵ=0.8$, $A=0.01{\mathrm{m}}^2$, $T=500\mathrm{K}$ and $T_s=300\mathrm{K}$. These are stipulated model inputs, not measured material properties. Using $\sigma \approx 5.670374419\times 10^{-8}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$ givesThe incoming irradiance is $459.300\mathrm{W}{\mathrm{m}}^{-2}$, corresponding to $4.5930\mathrm{W}$ incident on the surface; only $80\%$ is absorbed. At $T=T_s=300\mathrm{K}$, emitted and absorbed powers are both $3.6744\mathrm{W}$, so radiative equilibrium has zero net exchange despite continuing emission. Changing only emissivity to $0.2$ reduces both one-way powers and the net loss by a factor of four, giving $6.1694\mathrm{W}$ net. A low-emissivity coating therefore suppresses both radiative cooling to colder walls and radiative heating from hotter walls.

Near equal temperatures, write $T=T_s+\Delta T$. Taylor's theorem givesFor constant emissivity, the radiative heat transfer coefficient $h_{\mathrm{rad}}$ is approximately $4.8992\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$ at $T_s=300\mathrm{K}$ and $ϵ=0.8$. This local linear approximation is useful for small temperature differences; the $200\mathrm{K}$ difference in the worked example needs the full fourth-power expression.

A cooling experiment requires a wider first law of thermodynamics balance. In a uniform-temperature model without a heater,where $C(T)$ is the sample's heat capacity and the last two terms are signed outward convection and thermal conduction losses. A vacuum suppresses convection but does not eliminate thermal conduction through supports or sensor leads. A cooling slope therefore cannot establish emissivity unless these paths and heat capacity are independently constrained. Multiple wall temperatures, incomplete view factors, wavelength-selective materials or appreciable intervening absorption require a more complete radiative transfer calculation.

An optical concentrator redirects light from an entrance area $A_{\mathrm{in}}$ to a smaller receiver area $A_{\mathrm{out}}$. Its geometric concentration ratio is $C=A_{\mathrm{in}}/A_{\mathrm{out}}$. Specify the admitted angular distribution and losses before interpreting $C$ as an irradiance gain: if a fraction $\eta$ of incident radiant flux reaches the receiver, the ratio of mean receiver to mean entrance irradiance is $\eta C$.

A bidirectional reflectance distribution function, or BRDF, describes how a surface scatters incident illumination into each outgoing direction. At a fixed surface point $p$ and wavelength, its inputs are two directions above the surface, ${\omega }_i$ toward the incident source and ${\omega }_o$ toward the observer; equivalently these are two pairs of angular coordinates. Its output is the ratio of differential reflected radiance to differential incident irradiance:Position and wavelength may be additional arguments when material properties vary. A colour implementation can return one value per colour channel. The BRDF is a material response, not the final outgoing light value: it must be combined with the incident illumination and the projected-angle factor.^[ref]

For incident radiance $L_i$, the reflection part of the rendering equation isEmission is added separately if the surface itself emits. This integral explains how a BRDF replaces the assumed material response of a local lighting model. For Lambertian reflectance, $f_r={\rho }_d/\pi$, so the cosine factor yields the diffuse term from part (a). A directional specular lobe supplies a highlight. The empirical Phong reflection model is an approximate angular-response model of this kind; its original direct specular term is not automatically a normalized, energy-conserving BRDF. The ambient term approximates missing illumination and is not itself a third direction-independent material scattering mechanism. More general BRDFs can reproduce measured surface behaviour, while illumination from other surfaces supplies indirect light in the rendering equation.

At fixed spectrum and temperature, photocurrent is approximately proportional to irradiance, so the magnitude of short-circuit current increases nearly linearly. In the ideal model,so open-circuit voltage rises only logarithmically with irradiance. More incident optical power generally raises the maximum electrical power, within the device's thermal and resistive limits.

Increasing temperature strongly increases the dark saturation electric current. In silicon this normally reduces open-circuit voltage and maximum power, despite a small increase in short-circuit current associated with the temperature dependence of absorption and band gap. The thermal-voltage factor alone is insufficient to predict the sign; the temperature dependence of $I_0$ matters.

maximum power point tracking adjusts the load seen by the cell as irradiance and temperature change. A controlled DC–DC converter can transform the external load so that operation stays near the changing maximum of $-VI$, instead of at the generally suboptimal intersection with a fixed load. It matches operating conditions and cannot generate additional energy beyond the photovoltaic output.

The condenser lens should image the point source onto the small projection lens. Then every illuminated LCD point directs its central light ray into the projection lens, rather than wasting light elsewhere. In the paraxial idealization this gives approximately uniform useful illumination.

Put the LCD at $z=0$, the source at $z=-100\mathrm{mm}$, the projection lens at $z=u=29.5567\mathrm{mm}$, and the condenser lens at $z=-h$. The cone from the condenser lens's rim, of radius $15\mathrm{mm}$, to the projection lens has radiusat the LCD. Maximum approximately uniform illumination without unnecessary overfilling makes this equal the half-diagonal $7.5\mathrm{mm}$. Thus $h=u$: place the condenser lens $29.56\mathrm{mm}$ before the LCD, or $70.44\mathrm{mm}$ after the source. Its object and image distances are $100-h$ and $u+h$, soThe illuminated disk just contains the rectangular LCD. Exact spatially uniform irradiance is not guaranteed by a single spherical lens and a point source; the result uses paraxial geometrical optics, negligible optical aberration, and an approximately uniform incident cone.

The Stefan–Boltzmann law gives solar luminosity $L_{\odot }=4\pi R_{\odot }^2\sigma T_{\odot }^4$. At orbital distance $r$, the incident irradiance is $\sigma T_{\odot }^4{R}_{\odot }^2/r^2$. A spherical absorber intercepts a disc of area $\pi R_M^2$ but radiates over $4\pi R_M^2$. Thus the uniform-temperature radiative equilibrium estimate isThe planetary radius cancels. This is a black-body radiation estimate with zero reflected fraction and complete redistribution of absorbed heat. More generally the balance fixes $⟨T^4{⟩}^{1/4}$, rather than the arithmetic mean of a nonuniform surface temperature. It excludes atmospheric heating and thermal lag; no measured surface temperature is inferred.

Let the spacecraft radius be $a$, and assume that heat redistribution keeps its whole surface at a common temperature $T$. Neglect background radiation, and let $R\gg r,a$. The Sun's luminosity, from the Stefan–Boltzmann law, isAt distance $R$, the incident irradiance is $F_S=L_S/(4\pi R^2)=\sigma T_S^4(r/R{)}^2$. The spacecraft intercepts radiation over its projected disc, area $\pi a^2$, but emits from its whole spherical area $4\pi a^2$. The radiative equilibrium condition is thereforeCancelling the common factors givesThe radius of the spacecraft cancels. The factor $1/\sqrt{2}$ follows from the ratio of absorbing projected area to emitting surface area. Uniform surface temperature is the intended additional idealization behind a single equilibrium surface temperature. A poorly conducting spacecraft would have position-dependent temperatures; for example, a locally reradiating patch facing the Sun would instead have $T_{\mathrm{subsolar}}=T_S\sqrt{r/R}$, and a different maximum-temperature constraint.

Normalization and angular pattern. A scattering phase function describes the distribution of single-scattered power over outgoing solid angle, for a specified incident direction, wavelength and polarization convention. Use the common normalizationThe scattering asymmetry parameter is the mean cosine of the scattering angle: forward scattering has positive $g$, backward scattering negative $g$, and a forward–backward symmetric pattern has $g=0$.

Let $C=E_0(2\pi /\lambda {)}^4[(n^2-1)/(n^2+2){]}^2{r}^6$. The given small-sphere angular intensity is $I=C(1+{\mu }^2)/2$, where $\mu =\cos \Theta$. Its total scattered power isNormalizing givesDimensionally the supplied $I$ is radiant intensity, power per solid angle. If interpreted as detector irradiance at a distance, the appropriate inverse-square distance factor is also needed; it cancels out of the normalized angular pattern.

The polar plot has equal forward and backward lobes, each with radius $3/2$, and lateral values $3/4$. A small nonabsorbing particle behaves as an induced electric dipole. Radiation transverse to the dipole and the average over two orthogonal incident polarizations produce the $1+{\cos }^2\Theta$ form. Its odd angular moment vanishes:As the size parameter $2\pi r/\lambda$ grows, Rayleigh scattering in the electric-dipole approximation no longer suffices. Higher multipoles and interference produce structure in the angular pattern, and large particles generally develop a strong narrow forward diffraction peak. Their $g$ is normally positive, although detailed values and angular features depend on refractive index, size distribution and shape; they are not fixed by radius alone. Large spheres may show rainbows and backscattering structure, while nonspherical ice introduces further angular features.

Optically thick transmission. In a nonabsorbing two-stream or transport approximation, it is the transport optical depth $(1-g)\tau$ that controls the loss of forward directional memory. Using the same idealized albedo closure as in option 1 gives total forward-hemisphere transmittanceFor the same large but finite extinction optical depth,The respective transport-depth factors are $0.15$, $0.30$ and $1$. The water cloud scatters most strongly forward and returns the least light backward. All transmittances tend to zero in the infinite-depth limit. The strictly unscattered direct-beam transmittance is instead $e^{-\tau }$, identical in the three cases: the ranking concerns total transmitted radiation, including forward-scattered light, not that direct component alone. Full radiative transfer can also depend on higher angular moments beyond $g$.

Write $\kappa =R_d/c_p$, where $R_d$ is the specific gas constant and $c_p$ the specific heat capacity at constant pressure. Conservation of potential temperature on a dry adiabatic process givesLet $Q=(1-{\alpha }_c)S$ be the absorbed stellar irradiance, and let $T_0=(Q/\sigma {)}^{1/4}$, with $\sigma$ the Stefan-Boltzmann constant. The transparent gas contributes no thermal radiation. For a cloud with emissivity $e$, Kirchhoff's law of thermal radiation gives thermal absorptivity $e$; assume thermal transmissivity $1-e$ and negligible thermal reflection, as in the usual one-layer grey body model. The outgoing radiant flux is the transmitted surface emission plus upward cloud emission:ThusNo factor of four is needed: $S$ is the incident irradiance of the specified column, rather than a stellar-beam irradiance that still needs a spherical average.

For fixed ${\alpha }_c,p_c,p_s$, increasing $e$ monotonically warms the ground when $0<q<1$. At $e=0$, $T_g=T_0$; at $e=1$, $T_c=T_0$ and $T_g=T_0/q$. A cloud at the surface, $q=1$, produces no such warming. Comparing with a genuinely cloud-free column also requires including its different albedo: relative to $(S/\sigma {)}^{1/4}$, the ratio is $[(1-{\alpha }_c)/(1-e+e{q}^4){]}^{1/4}$. Reflection can outweigh the greenhouse effect.

The governing balance here is the whole-column radiation budget. Imposing an additional purely radiative surface balance would incorrectly remove the convective heat transfer that the prescribed dry adiabatic process may require.

The habitable zone is the range of orbital distances where an appropriate planetary atmosphere could sustain liquid water at the surface; it is not a guarantee of inhabited or even habitable conditions for every planet. At a fixed habitable incident flux, $a_{\mathrm{HZ}}\propto L_{\star }^{1/2}$. With the supplied $L_{\star }\propto M_{\star }^3$, Kepler's third law givesTaking Earth as a rough central reference, a $0.1M_{\odot }$ star has $a_{\mathrm{HZ}}\sim 0.0316\mathrm{AU}$ and $P_{\mathrm{HZ}}\sim 6.5$ days. Its six-day, Earth-sized companion receives about $1.11$ times terrestrial irradiance and could therefore lie in the habitable zone in this simplified model. Atmosphere, stellar activity, tidal locking and volatile retention remain unknown. The six-day companion around the Sun-like star receives about $239$ times terrestrial irradiance and is approximately Jupiter-sized; it is not a plausible surface-water Earth analogue. The stellar spectral dependence of actual habitable-zone edges is omitted by the simple scaling.

The photoelectric saturation current is the photocurrent reached when increasing the collecting electric potential difference no longer increases the fraction of available emitted electrons collected. It is saturation with collecting voltage at fixed illumination. It does not mean that increasing incident light intensity cannot increase emission.

If $P$ is incident radiant flux on the emitting area and $\eta$ is the external quantum efficiency for producing escaping electrons, then the incident photon arrival rate is $P/(h\nu )$ and complete collection gives $I_{\mathrm{sat}}=e\eta P/(h\nu )$. For the example's $\nu =8.00\times 10^{14}\mathrm{Hz}$, choose $\eta =0.10$ and $P=1.00\mu \mathrm{W}$. The calculated photon arrival rate is $1.886488\times 10^{12}{\mathrm{s}}^{-1}$ and $I_{\mathrm{sat}}=30.224866\mathrm{nA}$. Doubling $P$ doubles this current to $60.449731\mathrm{nA}$ while leaving $V_s$ fixed. At fixed illuminated area this doubles the irradiance. Changing frequency at fixed $P$ changes the arrival rate as well as the photon energy; $\eta$ can also change, so there is no general rule that a higher frequency gives a larger photocurrent.

A missing plateau can indicate incomplete collection or an electric field altered by the emitted electric charge. Heating, surface changes and additional emission channels can invalidate a constant $\eta$. Even a small amount of unwanted higher-frequency radiation can move an apparent endpoint. A stopping potential inferred from zero net current additionally needs background subtraction and a check for emission from the collector itself.

Radiance resolves both position and direction: it is radiant flux per projected area and per solid angle. For a surface element of area $dA$, and a small outgoing cone $d\Omega$ at angle $0\le \theta <\pi /2$ to its outward unit normal,Here $\mathbf{x}$ locates the patch and $\mathbf{s}$ is a unit vector along the outgoing direction. The factor $dA\cos \theta$ is the patch's apparent area viewed along that direction. The same local definition applies to transmitted or received radiation, choosing the positive projected area for the crossing direction. ^[ref].

For a planar patch of area $A$ with spatially uniform, direction-independent radiance $L$, integrating over the patch givesTake $L=10\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$ and $A=0.020{\mathrm{m}}^2$. The radiant intensities are $I(0)=0.20\mathrm{W}\mathrm{s}{\mathrm{r}}^{-1}$ and $I(60^{\circ })=0.10\mathrm{W}\mathrm{s}{\mathrm{r}}^{-1}$. A sufficiently narrow $0.010\mathrm{sr}$ cone therefore carries approximately $2.0\mathrm{mW}$ or $1.0\mathrm{mW}$, respectively. Both views have the same radiance $L$: the oblique view presents half the projected area. This is Lambert's cosine law, not a loss of radiance with viewing angle.

Radiance is useful because it separates a source's local directional output from the geometry of an observation. Integrating over outgoing directions gives radiant exitance; integrating incoming radiance at a receiver gives irradiance. In the stationary vacuum setting of vacuum conservation of specific intensity, a resolved uniform source retains its radiance along each light ray, while the radiant flux received from the whole source can decrease as its apparent solid angle shrinks.

Radiant exitance is total outward emitted radiant flux per actual emitting area, measured in $\mathrm{W}{\mathrm{m}}^{-2}$. Integrate radiance over the outward hemisphere, keeping the projected area factor:If the radiance is direction-independent, the spherical coordinates element is $d\Omega =\sin \theta d\theta d\phi$, soThe weighted solid angle is $\pi \mathrm{sr}$, although the whole hemisphere covers $2\pi \mathrm{sr}$. Directions near grazing contribute less because of the projected area. This geometric factor is the bridge between Lambert's cosine law and the total output of a diffuse patch. ^[ref].

For the uniform patch with $L=10\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$, radiant exitance is $M=10\pi \mathrm{W}{\mathrm{m}}^{-2}$. Its area $0.020{\mathrm{m}}^2$ gives total radiant flux $\Phi =MA=0.20\pi \mathrm{W}\approx 0.628319\mathrm{W}$ over the outward hemisphere. This total differs from its radiant intensity $I(0)=0.20\mathrm{W}\mathrm{s}{\mathrm{r}}^{-1}$, which describes only one direction.

For an ideal blackbody, integrate Planck's law over frequency as well: $M=M_e=j^{\ast }=\sigma T^4$ by the Stefan–Boltzmann law. An ideal $6000\mathrm{K}$ emitting patch of area $1\mathrm{m}{\mathrm{m}}^2$ emits about $73.4881\mathrm{W}$ outward. Irradiance counts incoming radiation separately; for purely emitted output the net outward exchange per area is $M-E$. Reflected output must also be included if the surface reflects incident radiation.

Radiant flux is the power carried by electromagnetic radiation: radiation energy transferred per unit time, measured in watts, $\mathrm{W}=\mathrm{J}{\mathrm{s}}^{-1}$. Specify what is counted: all radiation emitted by a source, radiation within one beam, or radiation intercepted by a photodetector. Writing $Q(t)$ for the cumulative energy crossing the specified boundary givesThe first expression is an instantaneous derivative; the second is an average over the stated time interval. If $0.020\mathrm{J}$ arrives during $0.010\mathrm{s}$, the mean radiant flux is $2\mathrm{W}$. The photodetector's absorbed power equals this incident radiant flux only if it absorbs all of the incident radiation.

Keeping radiant flux fixed while widening a beam reduces its irradiance, the incident power per receiving area. Radiance additionally resolves the propagation direction and has units $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$; an optical camera can collect different radiant fluxes from sources with the same radiance but different apparent solid angles. Luminous flux, measured in lumens, weights the optical spectrum by the human eye's standardized response instead of counting energy equally at every wavelength. Equal radiant fluxes need not give equal luminous fluxes.^[ref]

Radiant power also differs from the optical power of an optical lens, which is reciprocal focal length in dioptres. Likewise, astronomical radiative flux is power per area. Stating the units prevents these shared words from hiding different physical quantities.

In geometrical optics, suppose radiation has a nonnegative radiance $L(\mathbf{x},\mathbf{s})$ whose directional contributions add as power. Here $\mathbf{s}$ is the actual propagation unit vector, $d\Omega$ is a solid angle element, and the collecting oriented surface $A$ has chosen unit normal $\mathbf{n}(\mathbf{x})$. Assume the following integrals exist. The outward and inward radiant fluxes areBoth quantities are nonnegative; their difference is the signed net radiant flux. The dot product supplies the cosine of the angle with the unit normal, hence the projected area. For incoming radiation the outward unit normal gives a negative dot product; the minus sign makes the collected power positive. Measuring the incoming angle from a source-facing unit normal instead gives the familiar positive $\cos \theta$. The units check is $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\times {\mathrm{m}}^2\times \mathrm{sr}=\mathrm{W}$.

Equivalently, the incoming surface integral is ${\Phi }_{-}={\int }_{A}EdA$ with local irradiance $E$; the emitted part of ${\Phi }_{+}$ is ${\int }_{A}MdA$ with radiant exitance $M$. For a small flat patch and a narrow accepted cone, nearly constant radiance and angle give $\Phi \simeq LA\cos \theta \Delta \Omega$. This approximation requires uniformity across the patch and cone. Direction-independent outgoing radiance over a whole hemisphere gives the different exact result ${\Phi }_{+}=\pi LA$, by Lambert's cosine law and the spherical coordinates calculation in radiant exitance. Replacing the cosine-weighted integral by the hemisphere's $2\pi$ solid angle would double the answer.^[ref]

These radiance formulas presume a description in additive directional power; they do not reconstruct arbitrary coherent wave interference from separate ray contributions. For general electromagnetic fields, the signed instantaneous power crossing an oriented surface is ${\int }_A\mathbf{S}\cdot \mathbf{n}dA$, using the Poynting vector $\mathbf{S}$; optical measurements commonly average over many wave periods. A zero net radiant flux can conceal equal nonzero incoming and outgoing radiant fluxes.

For normal incidence in vacuum, a uniform beam delivers energy $IA\Delta t$ to an illuminated surface area $A$ during time $\Delta t$. A perfect absorber receives momentum $IA\Delta t/c$, so force divided by surface area gives $p_{\mathrm{abs}}=I/c$. A perfect stationary mirror sends the photons back with reversed momentum, so momentum conservation gives twice the normal transfer, $p_{\mathrm{refl}}=2I/c$. These are time-averaged pressures; the microscopic arrivals need not be uniform.

For a stipulated model light intensity $I=1000\mathrm{W}{\mathrm{m}}^{-2}$ and the exact SI speed of light $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$,These are computed predictions, not measured data or a universal sunlight value. If the beam fully covers a normally oriented surface area $A=0.010{\mathrm{m}}^2$, the intercepted radiant flux is $\Phi =IA=10\mathrm{W}$ and the absorber's force is $\Phi /c\simeq 33.356\mathrm{nN}$. Using the total source power would be wrong if only part of its beam reaches the receiver. The absorption prediction assumes complete photon absorption, while the reflection prediction assumes complete specular reflection; transmission, scattering of light and directed thermal photon emission require their own outgoing momentum accounting.

Measure the angle of incidence $\theta$ between the incoming path and the surface's normal line, with $0\le \theta <\pi /2$. There are two separate cosine factors. A fixed actual surface area $A$ intercepts only its projected area $A\cos \theta$, so $\Phi =IA\cos \theta$. Each incoming photon supplies normal momentum $(E_{\gamma }/c)\cos \theta$ on absorption, or twice that on specular reflection. Thus, for radiometric absorptance $a$ and reflected power fraction $R$,This model assumes transmitted light continues straight in the same medium and carries away its original momentum. It omits thermal recoil and assumes a stationary receiver. At $\theta =60^{\circ }$, fixed beam light intensity therefore gives one quarter of the normal-incidence pressure. For fixed intercepted radiant flux $\Phi$, however, the normal force has only one cosine factor. Equivalently, if the given quantity is receiving-surface irradiance $E_{\mathrm{inc}}=I\cos \theta$, then $p_{\perp }=(a+2R)E_{\mathrm{inc}}\cos \theta /c$. The distinction matters when interpreting an experiment's angle scan. ^[ref]

Absorption also transfers a tangential momentum component at oblique incidence. It contributes a tangential force, rather than normal pressure. Ideal specular reflection preserves the tangential momentum component, so it transfers only normal momentum.

A radiative-convective climate column applies a radiative-convective equilibrium model to the vertical thermal structure relevant to climate change. Radiative transfer determines heating and cooling; convective adjustment represents unresolved vertical redistribution when the chosen atmospheric lapse rate is too steep. Radiative equilibrium alone need not give a physically admissible lower-atmospheric profile.

In 1967, Syukuro Manabe and Richard T. Wetherald computed equilibrium with prescribed relative humidity, allowing atmospheric water vapor to change with temperature. Their average-cloud case warmed by $2.36\mathrm{K}$ for carbon dioxide increasing from 300 to 600 parts per million, versus $1.33\mathrm{K}$ with fixed absolute humidity. These are model outputs, not measured warming. The comparison isolates an important water-vapor feedback.^[ref]

An original two-temperature calculation exposes the bookkeeping. Let an opaque surface absorb solar irradiance $Q$ and emit blackbody thermal radiation $\sigma T_s^4$. A single isothermal atmospheric layer transmits all sunlight and absorbs a fraction $ϵ$ of surface infrared emission, emitting $ϵ\sigma T_a^4$ both upward and downward. This is a grey thermal-band emissivity, with $0<ϵ\le 1$. Let $H$ be upward nonradiative heat flux, and let $C_s,C_a>0$ be heat capacities per unit area. Then conservation of energy givesThe two equations sum toHere $L$ is outgoing longwave radiation. Both $H$ and downward atmospheric emission cancel internally: they redistribute energy without creating it. The factor two in atmospheric emission counts two faces. Dropping the downward face would spoil the surface budget. Each right-hand term has units $\mathrm{W}{\mathrm{m}}^{-2}$; each $C$ has units $\mathrm{J}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$.

Choose $Q=240\mathrm{W}{\mathrm{m}}^{-2}$, $H=80\mathrm{W}{\mathrm{m}}^{-2}$ and $ϵ=1$. At equilibrium, $L=\sigma T_a^4=240\mathrm{W}{\mathrm{m}}^{-2}$ and $\sigma T_s^4=400\mathrm{W}{\mathrm{m}}^{-2}$, giving $T_a\simeq 255.06\mathrm{K}$ and $T_s\simeq 289.81\mathrm{K}$. Direct substitution gives surface balance $240+240-400-80=0$ and atmospheric balance $400-480+80=0$. These deliberately chosen fluxes illustrate the model; they are not an observed global budget or a reconstruction of the 1967 calculation. Prescribing $H$ illustrates heat transfer accounting but does not close a realistic convection law.

For fixed $Q,H,ϵ$, the positive-temperature equilibrium is locally stable. Put $a=4\sigma T_s^3>0$ and $b=4ϵ\sigma T_a^3>0$. The Jacobian matrix of the temperature equations has trace $-a/C_s-2b/C_a<0$ and determinant $ab(2-ϵ)/(C_s{C}_a)>0$. Thus both eigenvalues have negative real part. Temperature-dependent climate feedbacks alter this test. A single grey layer also omits the optical spectrum, cloud structure and the independent thermal behavior of the troposphere and stratosphere.

A black surface receiving absorbed stellar irradiance $Q$, beneath a single non-solar-absorbing cloud with thermal emissivity $e>0$, satisfies $T_c^4=T_g^4/2$ when both bodies are in pure radiative equilibrium. The surface balance is $\sigma T_g^4=Q+e\sigma T_c^4$, giving the displayed result. If instead convection pins $T_c=q{T}_g$, the whole-column balance gives $T_g^4=Q/[\sigma (1-e+e{q}^4)]$; the nonradiative energy budget cannot then be discarded.

Radiometry measures the energy and power carried by electromagnetic radiation, including light. It asks how much radiation a source supplies, where it travels, and how much reaches a receiving surface. The measurements are physical rather than weighted by the sensitivity of the human eye: equal radiant fluxes at different wavelengths need not look equally bright. ^[ref].

The five quantities below answer different questions about the same radiation. The International System of Units makes the distinctions visible: a watt is a joule per second, while a steradian measures solid angle. ^[ref].

Resolving wavelength or frequency adds a further “per unit interval”, as in spectral radiance. The notation is consistent with NIST's table of spectral quantities. In particular, radiant flux is power, whereas the existing astronomical radiative flux is power per area. ^[ref].

This bookkeeping makes an optical camera's collected signal calculable from a source's radiance and the geometry of the receiving aperture. It also lets radiation measurements from different instruments be compared. For example, NIST describes separate calibrations of spectral radiance and spectral irradiance for instruments used to observe Earth. ^[ref].