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Black-body radiation is the electromagnetic radiation in thermal equilibrium with an ideal blackbody. It can be described as a photon gas: a cavity whose walls are all at the same temperature contains radiation with the universal Planck's law spectrum. A small opening in such a cavity approximates a blackbody emitter because incoming radiation undergoes repeated absorption inside. The radiation's equilibrium spectrum depends on temperature, rather than the wall material. ^[ref]

There are three useful quantities to keep distinct: directional spectral radiance $B_{\nu }$ or $B_{\lambda }$, total energy density $u$ in the cavity, and outward radiant exitance $j^{\ast }$ from its opening. Planck's law supplies the first; integrals over frequency and directions supply the others. For isotropy, every direction has the same $B_{\nu }$, so the full $4\pi$ solid angle givesHere $\hslash =h/(2\pi )$ is the reduced Planck constant, and the radiation constant is $a_{\mathrm{rad}}\approx 7.565733250\times 10^{-16}\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-4}$. The explicit Boltzmann constant makes this an SI formula; adopting natural units with $k_B=1$ recovers $u={\pi }^2{T}^4/(15{\hslash }^3{c}^3)$.

For a short integral outline, substitute $x=h\nu /(k_{B}T)$ and useThe expansion is a geometric series with nonnegative terms; repeated integration by parts evaluates each integral, and $\zeta (4)={\pi }^4/90$ is a value of the Riemann zeta function. This gives the coefficient above.

The photon gas also exerts radiation pressure, $p=u/3$. The factor $1/3$ comes from the angular average of the squared directional component of photon momentum, $⟨{\cos }^2\theta ⟩=1/3$. The thermodynamic derivation of blackbody radiation pressure gives the same result. In contrast, outgoing power involves one projected-direction factor over an outward hemisphere, yielding radiant exitanceThis is the Stefan–Boltzmann law with unit emissivity; $\sigma$ is the Stefan-Boltzmann constant. The factor $\pi$ follows from Lambert's cosine law, rather than the $4\pi$ full-sphere factor used for energy density.

For example, an ideal $6000\mathrm{K}$ cavity has $u\approx 0.980519\mathrm{J}{\mathrm{m}}^{-3}$, radiation pressure $p\approx 0.326840\mathrm{Pa}$, and outward radiant exitance $j^{\ast }\approx 7.34881\times 10^7\mathrm{W}{\mathrm{m}}^{-2}$. A stipulated $3000\mathrm{K}$ cavity has one-sixteenth each value because all three scale as $T^4$. These describe stored energy, mechanical pressure, and outgoing power per area, respectively. In a closed equilibrium cavity, equal opposing radiation balances the directional power flow even though $u$ and $p$ are nonzero.

The usefulness of black-body radiation is this common reference: Planck's law, Wien's displacement law, and the Stefan–Boltzmann law consistently connect spectral measurements, temperature, and total emission. The three illustrative curves in Planck's law are ideal equilibrium spectra; applying them to a material requires considering its emissivity and whether thermal equilibrium is a suitable approximation.

The adjustment from an astronomical magnitude in a particular observational band to the bolometric magnitude, accounting for energy outside that band. It depends on the stellar spectrum, hence on effective temperature, composition and surface gravity. A ratio of observed single-band luminosities should not be inserted into the Stefan–Boltzmann law as a bolometric ratio without allowing for different corrections.

For gas-pressure-supported stellar homology with specific nuclear heating $ϵ={ϵ}_0\rho T^{13}$ and opacity $\kappa ={\kappa }_0\rho T^{-3}$, the dimensionless radiative coefficient is proportional to $L{\mu }^{-7}{M}^{-5}$. The nuclear coefficient is proportional to ${\mu }^{13}{M}^{15}/(L{R}^{16})$. Holding both and the dimensionless profiles fixed gives the displayed scalings. The Stefan–Boltzmann law then gives $T_e\propto {\mu }^{25/16}{M}^{15/16}$ and $L\propto {\mu }^{-4/3}{T}_e^{16/3}$. The coefficients, pressure regime and composition assumptions must be held as specified; these relations are not universal laws for massive stars.

Oberlin College's documented demonstration heats a cube internally with a bulb and compares polished aluminium, dull aluminium, black-painted and white-painted faces at the same temperature. The apparatus includes a thermopile and a digital meter. Its preparation specifies 20 minutes of preheating and heater setting 6; these are instructions, not measured outcomes. The thermopile supplies the electrical readout for the relative thermal radiation comparison. Cover it between readings: otherwise detector warming changes its reference temperature. The account reports little difference between painted black and white faces and a larger contrast between shiny and painted faces. The comparison motivates a surface-dependent emissivity and shows why visible color alone is insufficient.^[ref]

Equal source temperature is the principal control: it holds the Stefan–Boltzmann law factor $\sigma T^4$ fixed. For a sharper comparison, also hold detector distance, viewing angle, viewed area and surrounding temperatures fixed, and record a covered-detector baseline. These are additional diagnostic controls. Under those assumptions, a changed readout can reveal changed directional emissivity averaged over the detector's wavelength response. The demonstration supplies neither an absolute calibration nor a scan over all directions and wavelengths; it therefore does not directly measure total hemispherical emissivity or establish the fourth-power law. Comparing cooling rates would additionally require separating convection, thermal conduction and heat capacity effects, as in net radiative heat exchange.

For a fixed-mass, fully convective monatomic gas protostar, write $P=K{T}^{5/2}$ and $K=K_0{\mu }^{-5/2}{M}^{-1/2}{R}^{-3/2}$. Matching the photosphere to $\kappa P=2g/3$ with $\kappa ={\kappa }_0\rho T^4$ gives $T_e^8=2G\mathcal{R}{\mu }^4{M}^{2}R/(3{\kappa }_0{K}_0^2)$. The Stefan–Boltzmann law gives $L=\pi ac[2G\mathcal{R}/(3{\kappa }_0){]}^{1/2}{K}_0^{-1}{\mu }^{2}M{R}^{5/2}$. The stellar virial theorem gives total energy $-3G{M}^2/(7R)$, so $L=-3G{M}^2\dot{R}/(7R^2)$. Integrating proves $C=(49\pi ac/6)[2\mathcal{R}/(3{\kappa }_{0}G){]}^{1/2}{K}_0^{-1}$. This holds before stellar nuclear fusion, changing opacity or electron degeneracy pressure invalidates the assumed contraction law.

For a monatomic fully convective ideal gas star, uniform specific entropy gives $P=K{T}^{5/2}$ with $K\propto {\mu }^{-5/2}{M}^{-1/2}{R}^{-3/2}$. Combining this with the stellar surface boundary condition $\kappa P=2g/3$ and $\kappa ={\kappa }_0{\rho }^a{T}^b$ givesAll composition comparisons here hold ${\kappa }_0$ fixed. For $a=1,b=4$, $T_e^8\propto {\mu }^4{M}^{2}R$, and the Stefan–Boltzmann law gives $L\propto {\mu }^{2}M{R}^{5/2}$. The stellar virial theorem and Kelvin-Helmholtz contraction then give $R^{-7/2}-R_0^{-7/2}\propto {\mu }^2{M}^{-1}(t-t_0)$, with a positive coefficient. Once the initial-radius term is negligible, $R\propto {\mu }^{-4/7}{M}^{2/7}(t-t_0{)}^{-2/7}$.

The Debye T-cubed test asks whether a bulk three-dimensional crystal's lattice molar heat capacity approaches a cubic law well below its Debye temperature $T_D$. It tests the low-frequency acoustic phonon branches in the Debye model, rather than asserting that every contribution to a real sample's heat capacity is cubic. Peter Debye's continuum description makes a striking connection: a thermal measurement can reveal the low-frequency vibrational density of states without resolving individual vibrations.^[ref]

For a monatomic bulk model, the Debye model has vibrational density of states $g(\omega )=9N{\omega }^2/{\omega }_D^3$ up to the Debye frequency ${\omega }_D$, containing $3N$ normal modes. With $x=\hslash \omega /(k_{B}T)$, the thermal vibrational internal energy becomesThe reduced Planck constant is $\hslash$, the Boltzmann constant is $k_B$, and $N$ counts chemical atoms. For $T\ll T_D$, replace the upper limit by infinity. The integral is ${\pi }^4/15$, as in the Stefan–Boltzmann law derivation. Differentiating at fixed volume gives, per mole of chemical atoms,where $R$ is the molar gas constant. The cubic exponent comes from three-dimensional acoustic phonon branches with linear low-wavevector dispersion relations. The coefficient encodes their propagation speeds. Formula-unit normalization in a compound requires its number of chemical atoms per formula unit; do not silently substitute moles of formula units for moles of chemical atoms.

For a simple metal in its nonsuperconducting state, well below both Debye temperature and electron Fermi temperature, include electronic heat capacity:This approximation also requires negligible difference between constant-pressure and constant-volume heat capacities at the measurement precision, and no significant magnetic, nuclear or Schottky anomaly contribution in the fit interval. Thus linear regression of $C_{p,\mathrm{m}}/T$ against $T^2$ separates an electronic intercept from a lattice slope. For an insulating crystal with no relevant low-energy magnetic excitations, the electronic term can instead be negligible; a superconductivity transition changes the electronic response.^[ref]

After subtracting calorimeter addenda, Brando's second gold sample, $72.6\mathrm{mg}$, gave a linear $C_{p,\mathrm{m}}/T$ versus $T^2$ plot in the nonsuperconducting state. The reported electronic intercept was $7.0\times 10^{-4}\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-2}$; deviations from reference heat capacities were below $2\%$.^[ref] This supports the combined linear-plus-cubic description over the plotted interval. Because the platform subtraction used reference gold, it is also an instrument validation with a shared material standard, rather than a wholly independent determination of the Debye model from an uncalibrated instrument.

For a worked original inference, stipulate values of $y=C_{p,\mathrm{m}}/T$ equal to $(0.001100,0.002300,0.004300)\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-2}$ at $T=(1,2,3)\mathrm{K}$. Assign independent $u(y)=0.000020\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-2}$ and negligible temperature error. Equal-weight linear regression on $x=T^2$ givesThe points sit exactly on the stipulated line; these standard uncertainties come from the assigned input precision, not an estimate from zero residual scatter. With $R=8.314462618\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, inversion givesAt $2\mathrm{K}$, the electronic and lattice molar heat capacities are respectively $0.0014$ and $0.0032\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. Their equality occurs at $\sqrt{\gamma /\beta }=1.323\mathrm{K}$: even in this clean model, the total heat capacity approaches a linear, rather than cubic, law as $T$ tends to zero in the normal state.

A straight line over a narrow interval is a useful diagnostic, not a universal proof. Inspect residuals and repeat the fit over several upper and lower temperature cutoffs; curvature can signal higher lattice terms, an unremoved Schottky anomaly, or a faulty calorimeter addenda subtraction. Correlated thermometer calibration and sample-amount errors affect both axes and can dominate the illustrative linear regression uncertainty. Low-dimensional samples, additional low-energy excitations and proximity to a phase transition need other models. The test is valuable precisely because its failures can reveal physics beyond the bulk acoustic approximation.

A Keplerian accretion disk with a zero-torque inner boundary condition and local blackbody cooling has $T_{\mathrm{eff}}^4=3GM\dot{M}(1-\sqrt{r_{\mathrm{in}}/r})/(8\pi \sigma r^3)$. The Stefan–Boltzmann law uses the flux through one face; the standard thin-disk dissipation flux includes both faces. The scale $T_{\mathrm{in}}=[3GM\dot{M}/(8\pi \sigma r_{\mathrm{in}}^3){]}^{1/4}$ is a normalization, not the actual temperature at the stress-free inner edge.

Assume a passive disk has local isothermal sound speed squared $c_s^2=aT$, disk scale height $H=c_s/{\Omega }_K$, and irradiated height $z_s=fH$ with constant $f$. Its irradiation grazing angle of a disk and Stefan–Boltzmann law give $q^{\prime }=C{q}^8/R^3$ for $q=z_s/R$, where $C=4\pi \sigma (GM/a{)}^4/(L_{\ast }{f}^8)$. Integrating gives $q^{-7}=R_0^2/R^2+B$, with $R_0^2=7C/2$. Where the first term dominates and the thin disk assumptions hold, $q\simeq (R/R_0{)}^{2/7}$, $H\propto R^{9/7}$ and $T\propto R^{-3/7}$. The law cannot be extrapolated into a geometrically thick region using the same grazing-angle approximation.

With negligible heavy-element contribution to the mean molecular weight, $\mu =4/(3+5X)$ and $\kappa \propto 1+X$. At fixed mass, composition-dependent CNO homology gives $L\propto (1+X{)}^{-1}(3+5X{)}^{-4}$ and $R\propto [X(1+X){]}^{1/16}(3+5X{)}^{-9/16}$. The Stefan–Boltzmann law then gives $T_e\propto X^{-1/32}(1+X{)}^{-9/32}(3+5X{)}^{-23/32}$. From the formal $X=1$ reference, declining $X$ initially increases luminosity and effective temperature, unlike the usual redward main-sequence turn-off. A trace CNO catalyst abundance is implicit in the burning coefficient; an exactly metal-free pure-hydrogen star cannot burn through that cycle.

Match a fully convective stellar polytrope of index $3/2$ to the grey-atmosphere convection onset with eighth-power temperature opacity. At the fixed matching optical depth, $T_b\propto T_e$ and $P_b\propto g^{2/3}{T}_e^{-16/3}$. The ideal-gas adiabat $P=K{\rho }^{5/3}$ gives $K\propto T_b^{5/3}{P}_b^{-2/3}\propto T_e^{47/9}{g}^{-4/9}$. The polytropic mass-radius relation gives $K\propto R{M}^{1/3}$; with $g\propto M/R^2$, this yields $T_e^{47}\propto R{M}^7$. The Stefan–Boltzmann law then gives the displayed luminosity. Composition and opacity normalization are held fixed between models.

Match an efficient, monatomic gas, fully convective star to the grey-atmosphere convection onset with thirteenth-power opacity. Its convective relation is $P=K_T{T}^{5/2}$. At the matching temperature $T_b\propto T_e$, the stellar atmosphere gives $K_T^2\propto M{L}^{-1}{T}_e^{-13}\propto M{R}^{13/2}{L}^{-17/4}$ using the Stefan–Boltzmann law. The ideal-gas relation converts this to $P=K_{\rho }{\rho }^{5/3}$ with $K_{\rho }=(\mathcal{R}/\mu {)}^{5/3}{K}_T^{-2/3}$. The polytropic mass-radius relation at index $3/2$ gives $K_{\rho }\propto G{M}^{1/3}R$, hence $K_T^2\propto M^{-1}{R}^{-3}$. Equating the two expressions proves $L^{17/4}\propto M^2{R}^{19/2}$, or $L\propto M^{8/17}{R}^{38/17}$. The entropy coefficient is constant throughout the convective envelope, but can change between models.

A Hertzsprung-Russell diagram plots stellar luminosity against effective temperature or a calibrated spectral classification. Conventionally, high luminosity is upward and high temperature is to the left. The Stefan–Boltzmann law, $L=4\pi R_{\star }^2\sigma T_{\mathrm{eff}}^4$, explains why stars of equal temperature but different luminosities must have different stellar radii.

The main sequence crosses the diagram, giant stars occupy a luminous region, and white dwarfs lie at low luminosity despite their high temperatures. An evolutionary track follows one model star through this space; a population diagram instead combines many stars at once. A colour-magnitude diagram is the observational counterpart, requiring distance, extinction and spectral calibration for the conversion.

For a stellar homology sequence with $R\propto M^X$ and $L\propto M^Y$, the Stefan–Boltzmann law gives $T_{\mathrm{eff}}\propto M^{(Y-2X)/4}$. Eliminating mass gives the displayed Hertzsprung-Russell diagram slope when $Y\ne 2X$. The usual horizontal axis decreases in temperature toward the right, reversing the visual left-to-right sign. The relation is between logarithmic variables and depends on the specified structural assumptions.

Let $\mathcal{R}=k_B/({\mu }_m{m}_p)$ for constant mean molecular weight ${\mu }_m$ in proton-mass units, and let $\sigma$ be the Stefan–Boltzmann constant in the Stefan–Boltzmann law. The gas sound speed used here satisfies $c_s^2=\mathcal{R}T$, and $\Omega$ is the local orbital frequency. For the local opacity approximation $\kappa ={\kappa }_0{\rho }^{1/3}{T}^{10}$, a gas-supported optically thick alpha disk has $\rho \sim \Sigma /H$ and $\mathcal{R}T\sim {\Omega }^2{H}^2$. Integrated viscous heating is $F_{+}\sim \alpha \Sigma {\Omega }^3{H}^2$. Radiative diffusion gives $F_{-}\sim (\sigma /{\kappa }_0)T^{-6}{\Sigma }^{-4/3}{H}^{1/3}\sim (\sigma /{\kappa }_0){\mathcal{R}}^6{\Omega }^{-12}{\Sigma }^{-4/3}{H}^{-35/3}$. Equating the fluxes proves the thickness scaling. At fixed composition and $\alpha$, it gives $H/r\propto r^{53/82}{\Sigma }^{-7/41}$ and $c_s\propto r^{6/41}{\Sigma }^{-7/41}$ in Keplerian rotation. Radial powers therefore require a surface-density profile; they are not determined by opacity alone. The formula is local to the stated opacity regime and small disk aspect ratio.

An accretion disk forms when infalling material has enough angular momentum to orbit before it can reach the central object. In a thin disc dominated by a central mass $M$, the orbital motion is approximately Keplerian rotation:Vertical hydrostatic equilibrium gives the disk scale height $H\sim c_s/{\Omega }_K$; a geometrically thin disc has $H/R\ll 1$. The surface density is $\Sigma =\int \rho dz$, and a steady inward mass flow rate is $\dot{M}=-2\pi R\Sigma v_R>0$.

Moving inward requires losing specific angular momentum. Internal stresses transport angular momentum outward; much of the mass can then accrete while an outer fraction spreads to larger radius. Molecular viscosity alone is generally inadequate for rapid astrophysical accretion. Turbulent and magnetic stresses, including those generated by the magnetorotational instability, provide effective transport. A common parametrization is $\nu =\alpha c_{s}H$ for an effective kinematic viscosity, with $\alpha$ a dimensionless stress parameter.

For a thin Keplerian accretion disk, the evolution equation isThis viscous evolution of an accretion disk occurs on the viscous timescale $t_{\nu }\sim R^2/\nu$, much longer than an orbital period in a thin disc with moderate effective stress.

The orbital specific energy is $-GM/(2R)$, so inward motion releases binding energy. In a steady, optically thick, radiatively efficient disc with zero torque at its inner edge, the emergent flux from each face isThe Stefan–Boltzmann law identifies the local effective temperature. Far from the inner boundary, $T_{\mathrm{eff}}\propto R^{-3/4}$, so the emitted spectrum combines hotter inner and cooler outer annuli. Integrating both faces gives $L_{\mathrm{disc}}=GM\dot{M}/(2R_{\mathrm{in}})$ under these assumptions. This is not a universal spectrum or efficiency: optical depth, inner torque, energy advection and outflows can change the radiation. The essential connection is outward angular-momentum transport, inward accretion, and release of binding energy.

Use spectral radiant exitance per unit wavelength. The Planck law isA hotter blackbody radiates more at every wavelength, its wavelength peak moves to shorter wavelengths according to Wien's displacement law, and its integrated flux rises as $\sigma T^4$ by the Stefan–Boltzmann law. The wavelength convention matters: a spectrum plotted per unit frequency has a different peak transformation.

For planetary equilibrium temperature, assume an isotropic blackbody star, a circular orbit, negligible internal heating, unit surface thermal emissivity and efficient global heat redistribution so the surface has one temperature. The received stellar irradiance is $F_{\ast }=\sigma T_{\ast }^4(R_{\ast }/a{)}^2$. A planet of radius $R_p$ intercepts $\pi R_p^2{F}_{\ast }$ but radiates over $4\pi R_p^2$, giving the projected-area factor in planetary energy balanceWithout the uniform-temperature assumption, this calculation gives $⟨T^4{⟩}^{1/4}$, not a unique arithmetic mean of surface temperature.

With Bond albedo $A=0.05$, the absorbed fraction is $1-A$. Hencea decrease of approximately $3.87\mathrm{K}$. This albedo weights incident stellar light and does not imply thermal emissivity $0.95$.

Define the planetary effective emissivity by outgoing top-of-atmosphere flux $F_{\mathrm{out}}={ϵ}_{\mathrm{eff}}\sigma T_s^4$. Maintaining the same absorbed stellar power with $T_s=312\mathrm{K}$ givesThe greenhouse effect allows thermal radiation to escape from colder layers, so the outgoing flux is less than a black surface at $T_s$ would emit. Downward atmospheric emission contributes to the surface energy balance; this does not violate the top-of-atmosphere energy balance.^[ref]

A single-layer greenhouse model makes the distinction quantitative. Take a shortwave-transparent, isothermal layer with infrared absorptivity and emissivity ${ϵ}_a$, above a surface black in the thermal infrared. Writing $Q=(1-A)F_{\ast }/4$, its surface and atmospheric balances areFor ${ϵ}_a>0$, $T_a^4=T_s^4/2$. Radiation escaping to space is $(1-{ϵ}_a)\sigma T_s^4+{ϵ}_a\sigma T_a^4$, soThus $0.849$ is the planet's effective emissivity; $0.301$ is the layer's infrared emissivity within this particular model. They are not interchangeable. Real atmospheres require wavelength-dependent absorption, a vertical temperature profile and nonradiative heat transport; the measured temperatures alone do not uniquely retrieve an atmospheric optical profile.

The heat capacity per unit horizontal area is $C=\rho c_{p}d$: mass density times specific heat capacity times atmospheric depth. Thus $C\dot{T}$ is the rate of stored thermal energy per unit area. The solar constant $Q$ is the incident radiative flux on a surface normal to sunlight; the ratio of the Earth's intercepting disk area to its spherical area is $1/4$. The absorbed fraction is $1-a$, where albedo $a$ is the reflected fraction. The outgoing radiative flux is $\sigma \gamma T^4$, a Stefan–Boltzmann law with effective radiation factor $\gamma$ incorporating the simplified greenhouse effect. Every term therefore has units of power per area.

The albedo curve decreases smoothly, passes through $(T_0,1/2)$, and approaches $1/4$ at high temperature. At $T=0$ it is $(2+\tanh \mu )/4$, close to $3/4$ when $\mu$ is large. Its derivative isAn ice-covered planet reflects more sunlight than exposed land or ocean. Warming removes reflective ice, lowers albedo, and increases absorbed radiative flux: this is ice-albedo feedback. The parameter $\mu$ controls the sharpness of this transition; the formula is an idealized global response, rather than a local melting law.

Balances and meanings. The first equation is conservation of energy per unit area. Here $c$ is effective areal heat capacity, $Q$ is the solar constant, and $Q/4$ accounts for the ratio of the Earth's intercepting disc area to its total surface area. The albedo $a$ reflects a fraction of that input. The outgoing thermal radiation is the Stefan–Boltzmann law modified by $\gamma (p)$, an effective transmission factor for the greenhouse effect; it normally decreases as atmospheric carbon dioxide increases.

The albedo relaxes towards its temperature-dependent equilibrium on the ice-adjustment time $t_i$. The atmospheric mass inventory per unit pressure is $M_{{\mathrm{CO}}_2}A/(M_{a}g)$, using the molar mass ratio of carbon dioxide to air and hydrostatic atmospheric mass. Its sources and sinks are volcanic input $v$, removal by chemical weathering over land area $A_L$ at rate $W$, and the atmosphere-to-ocean exchange $h(p-p_s)$. The ocean equation is a molar mass conservation balance: ${\rho }_{{\mathrm{H}}_2\mathrm{O}}{V}_{\mathrm{oc}}$ is ocean water mass, division by $M_{{\mathrm{CO}}_2}$ converts the incoming carbon-dioxide mass to moles, and $bC$ removes dissolved carbon by burial/export. With $C$ in moles per kilogram, $b$ has units of kilograms of water-equivalent per year.

Times and reduction. Linearize the temperature equation at fixed $a,p$ about radiative equilibrium. Its exponential relaxation time isThe equality uses $\sigma \gamma T^4=Q(1-a)/4$. A stock-over-flux thermal time, sometimes used as a scale, is four times this, $0.381\mathrm{y}$; both describe a rapid thermal response relative to the other processes.

For pressure in atmospheres, write $I=M_{{\mathrm{CO}}_2}A10^5/(M_{a}g)$ for the atmospheric carbon-dioxide mass per atmosphere. The factor $10^5$ is essential because $h$ is given per atmosphere. The pressure and slow ocean exponential relaxation times areThus $t_T\ll t_p\ll t_i=10^4\mathrm{y}<t_C$. The leading quasi-steady approximation givesSubstitution in the two remaining equations yieldsThe coefficient of $\dot{a}$ remains $t_i$; the printed $t_T$ in the reduced equation is a typographical error. Eliminating a fast temperature variable does not replace the ice-response time. The ocean time here is the slow time after atmospheric balance cancels the exchange flux; holding pressure fixed after imposing a carbon-dependent $p_s$ would describe a different conditional relaxation.

The supplied volcanic input givesabout $41.1\%$ of the stated present value $2.0\times 10^{-3}$; that value is about $2.43C_0$.

Carbon-dioxide levels. Since $A_{L}W=3.0\times 10^{11}\mathrm{kg}{\mathrm{y}}^{-1}=v$, the atmospheric balance with $p_s=C/K$ givesThis uses the specified present ocean concentration, as requested, rather than replacing it by the model's smaller volcanic-only equilibrium.

After sustained additional input $v_i$, $C$ barely changes during the atmospheric adjustment. At this stageThis is an intermediate balance, approached over several centuries. On the slower ocean time,with pressure following $p\simeq C/K+v_i/h$. Since $C(0)<C_{\infty }$, pressure continues to rise slowly. Its eventual level in this stipulated constant-parameter model isThe two-reservoir carbon adjustment therefore has a rapid rise towards the frozen-ocean balance followed by a much longer increase. The figure solves the coupled linear pressure–ocean equations, retaining their small inventory correction to the reduced time scales.

Take homologous profiles and hold composition, hence mean molecular weight, fixed. The stellar mass conservation equation and stellar hydrostatic equation give ${\rho }_c\propto M/R^3$ and $P_c\propto G{M}^2/R^4$. With ideal gas pressure support, $T_c\propto M/R$. Integrating the specific nuclear rate through homologous profiles then givesThe radiative diffusion in a star equation gives $L_{\mathrm{rad}}\propto R{T}_c^4/({\kappa }_c{\rho }_c)$. Here ${\kappa }_c\propto {\rho }_c^{1/2}{T}_c^{-5/2}\propto M^{-2}R$, soEquality of generated and transported luminosity implies $M^7{R}^{-8}\propto M^5{R}^{-1}$, whenceThis is the radiative homology with density-half inverse-five-halves opacity scaling.

The Stefan–Boltzmann law defines the effective temperature by $L=4\pi R^2{\sigma }_{\mathrm{SB}}{T}_{\mathrm{eff}}^4$. Therefore $T_{\mathrm{eff}}\propto M^{29/28}$. In theoretical Hertzsprung-Russell diagram coordinates, the slope isThe usual graphical convention puts increasing temperature to the left, so the line descends as one moves right. If the horizontal coordinate is $-\log T_{\mathrm{eff}}$, its plotted slope is $-132/29$.