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Use a gas-pressure-supported, optically thick alpha disk, with $\mathcal{R}=k/({\mu }_m{m}_p)$ and positive parameters. Here $\sigma$ is the Stefan–Boltzmann constant in the Stefan–Boltzmann law, and $F_{+}$ and $F_{-}$ below are heating and cooling flux estimates per disk face. Hydrostatic equilibrium and the ideal gas relation give $\mathcal{R}T\sim {\Omega }^2{H}^2$, while $\rho \sim \Sigma /H$. Integrating viscous heating over one semi-thickness and estimating radiative diffusion giveHere the stipulated negative hydrogen ion opacity exponent is $\kappa ={\kappa }_0{\rho }^{1/3}{T}^{10}$, which is a local approximation distinct from other empirical opacity fits. Substituting $T\sim {\Omega }^2{H}^2/\mathcal{R}$ and equating heating and cooling givesDropping numerical constants of order one is essential here. This is the negative-hydrogen-opacity alpha-disk thickness scaling. Its angular thickness estimate isFor a central mass $M$, Keplerian rotation gives $\Omega =(GM/r^3{)}^{1/2}$, so $H/r\propto r^{53/82}\Sigma (r{)}^{-7/41}$. The isothermal sound speed $c_s^2=\mathcal{R}T$ isAn adiabatic sound speed differs by the constant factor $\sqrt{\gamma }$ if the adiabatic index is fixed. For $\Sigma \propto r^{-p}$, constant $\alpha$, and fixed composition, the disk aspect ratio grows as $r^{(53+14p)/82}$, so there is disk flaring when $p>-53/14$; $c_s\propto r^{(6+7p)/41}$. The problem supplies no numerical parameters or specified $\Sigma (r)$, so no unique numerical $H/r$ or unconditional radial power can be inferred.

If the additional standard closure of a constant inward accretion rate well outside a zero-torque edge is intended, $\bar{\nu }\sim \alpha \Omega H^2$ and $\bar{\nu }\Sigma \simeq \dot{M}/(3\pi )$ imply $\Sigma \propto (\alpha \Omega H^2{)}^{-1}$. Eliminating it gives $H^9\propto {\Omega }^{-38/3}$ at fixed $\dot{M},\alpha$, hence $H/r\propto r^{10/9}$ and $c_s\propto r^{11/18}$. These stronger powers use that extra steady-flow assumption. All these local scalings cease to apply if the thin disk or the stated opacity regime fails.

The geometric albedo $A_g$ compares the planet's reflected brightness at full phase with that of a perfectly reflecting Lambertian surface in the form of a flat disc of the same projected area, illuminated by the same incident flux. The Bond albedo $A_B$ is the fraction of total incident stellar power reflected in all directions and over all wavelengths. They need not be equal: the planetary phase integral and the incident spectral weighting connect them.

For a blackbody star, the incident flux at orbital distance $a$ isAssume negligible intrinsic luminosity, thermal emissivity one, and efficient day-night heat redistribution giving uniform emission over the whole planet. The absorbed and emitted powers areUsing the Stefan–Boltzmann law gives the planetary equilibrium temperatureFor a representative emitting area encoded by a stellar flux redistribution factor $f$, the equivalent expression iswith $f=1/4$ for global uniform emission and $1/2$ for uniform dayside-only emission. The Bond albedo, rather than the geometric albedo, belongs in the energy budget. Internal heat, wavelength-dependent emissivity and imperfect redistribution require corresponding modifications.

At the photosphere, use the stated boundary pressure and the effective temperature relation $L=4\pi R^2\sigma T_{\mathrm{eff}}^4$. The radiation constant obeys $a_{\mathrm{rad}}c=4\sigma$ by the Stefan–Boltzmann law. Substitution into the radiative temperature gradient givesThis uses the approximate stellar surface boundary condition at optical depth $2/3$; it is not a universal photospheric gradient for arbitrary atmospheres.

The Stefan–Boltzmann law defines the effective temperature through $L=4\pi R^2{\sigma }_{\mathrm{SB}}{T}_{\mathrm{eff}}^4$. The stellar homology relations give $R\propto M^{1/13}\propto L^{1/71}$, soThusIn a logarithmic Hertzsprung-Russell diagram, with $\log L$ plotted vertically against $\log T_{\mathrm{eff}}$,The usual horizontal axis increases in effective temperature toward the left, so the hotter, more luminous end lies toward the upper left. The quoted algebraic slope uses increasing $\log T_{\mathrm{eff}}$ as the coordinate.

Retain the same stellar homology and gas-pressure assumptions, but use the specified CNO cycle law $ϵ\propto \rho T^{16}$ and constant electron-scattering opacity. The nuclear luminosity now scales asThe radiative diffusion in a star scaling becomesEquating the two yields $R^{19}\propto M^{15}$. Therefore radiative homology with CNO burning and electron scattering givesSince $R\propto L^{5/19}$, the Stefan–Boltzmann law givesThis is a model comparison using the exponent 16 prescribed here. Actual massive stars may have appreciable radiation pressure and convective cores, which violate the assumptions of this purely radiative gas-pressure family.

A fully convective monatomic ideal gas is an adiabatic stellar polytrope with index $n=3/2$. Its polytropic mass-radius relation givesAt the base of a thin grey atmosphere, the stellar surface boundary condition and electron-scattering opacity giveThe ideal-gas equation evaluated on the convective adiabat givesCombining these relations,The Stefan–Boltzmann law then gives

At radius $r$, radiation of luminosity $L$ gives matter of flux-mean opacity $\kappa$ the outward accelerationEquating this with the gravitational acceleration $G{M}_{\mathrm{BH}}/r^2$ gives the Eddington luminosity

Let $L=f{L}_{\mathrm{Edd}}$ and let $\mathcal{M}(r)$ be the radiative force multiplier relative to the Compton or electron-scattering acceleration. A steady spherical wind obeys mass conservation ${\dot{M}}_w=4\pi r^2\rho v$ and the radial Euler momentum equationWhen gas pressure is negligible, radiation accelerates the wind locally if $f\mathcal{M}>1$. Electron scattering alone has $\mathcal{M}\simeq 1$, so it launches a wind only near or above the Eddington ratio $f=1$. Bound-free absorption, spectral lines, or dust can give $\mathcal{M}\gg 1$ and allow a sub-Eddington source to launch a wind, provided the gas is not so highly ionized that those extra opacities disappear. A wind that has already acquired sufficient kinetic energy may coast through a region where $f\mathcal{M}<1$, but such a region cannot launch a cold wind from rest.

Write the innermost stable circular orbit asEstimating the emitting area as $4\pi R_{\mathrm{ISCO}}^2$, the Stefan–Boltzmann law givesand henceThus $T_{\mathrm{BB}}\propto M_{\mathrm{BH}}^{-1/4}$ at fixed $f$ and opacity: a larger black hole radiates from an area growing as $M_{\mathrm{BH}}^2$, faster than its Eddington luminosity grows. Prograde black-hole spin moves the ISCO inward and raises the characteristic temperature, whereas retrograde spin moves it outward and lowers the temperature. Relativistic transfer and the radial temperature profile change the numerical coefficient, not these leading scalings.

A fully convective monatomic perfect-gas star follows an $n=3/2$ adiabatic stellar polytrope. At the photosphere, $\tau =2/3$ and hydrostatic equilibrium in optical depth givesFor an ideal gas on an adiabat,Evaluating this at the photosphere givesThe $n=3/2$ polytropic mass-radius relation is $R\propto K{M}^{-1/3}$, soFinally the Stefan–Boltzmann law givesThis is the specified Hayashi track.

Along the pp-chain homologous sequence, $R$ is constant and therefore $T_c\propto M$. Normalizing to the Sun gives the transition mass

For the CNO cycle law $ϵ\propto \rho T^{23/2}$, the nuclear scaling becomesThe opacity law is unchanged, so $L_{\mathrm{rad}}\propto M^{11/2}{R}^{-1/2}$. Equating them givesand then

The effective temperature follows from the Stefan–Boltzmann law $L=4\pi R^2\sigma T_e^4$. Thus the pp branch has $T_e\propto M^{11/8}$ and $L\propto T_e^4$, whereas the CNO branch hasOn a Hertzsprung-Russell diagram, both branches rise toward larger luminosity and, conventionally, leftward toward larger temperature. They join near $4M_{\odot }/3$; the CNO branch has the steeper $L$--$T_e$ logarithmic slope and lies to the cooler side of the extrapolated constant-radius branch at fixed luminosity.

If $X$ is the hydrogen mass fraction and $X_{\mathrm{CNO}}$ the combined carbon-nitrogen-oxygen mass fraction, the reaction-pair and catalyst abundances give approximatelyThe CNO nuclei are catalysts, so their abundance multiplies the CNO-cycle rate rather than being consumed by the completed cycle.

The upward arrows at the surface are reflected short-wave flux $\alpha F_r$, emitted long-wave flux $ϵ\sigma T_s^4$ from the Stefan–Boltzmann law, and, when positive, conductive heat arriving from the planetary ice shell. The downward arrow is the incident solar flux $F_r$. The planetary surface energy balance is thereforeThe relevant planetary albedo is the Bond albedo. In a steady one-dimensional shell, Fourier's law and $k(T)=a/T$ make the upward conductive fluxwhere $z$ points downward and $T(h)=T_m$. Substitution gives the requested implicit equation

Planck's law gives the spectral radiance of an ideal blackbody in thermal equilibrium at absolute temperature $T>0$. It predicts how much power is carried in each small frequency or wavelength interval, per projected area and per solid angle. The whole curve is determined by $T$: no material-specific parameter enters this ideal Planck function.

Writing $\nu$ for frequency in hertz and $\lambda$ for wavelength in metres, the two forms of the Planck function areHere $h$ is the Planck constant, $k_B$ the Boltzmann constant, and $c$ the speed of light in vacuum. Their exact values in the International System of Units are $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. ^[ref]

The units matter: $B_{\nu }$ is measured in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{H}{\mathrm{z}}^{-1}$, whereas $B_{\lambda }$ is measured in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$. Both are spectral radiance, rather than radiant exitance or energy density. To report $B_{\lambda }$ per micrometre, multiply its per-metre value by $10^{-6}$; a micrometre-wide interval is $10^{-6}$ metres wide.

For a computed example, at $T=6000\mathrm{K}$ and $\lambda =0.500\mu \mathrm{m}$, the exponent is $hc/(\lambda k_{B}T)\approx 4.79592$. Substitution into Planck's law gives $B_{\lambda }\approx 3.17569\times 10^{13}\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$, or $3.17569\times 10^7\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$ per micrometre. At the same wavelength, the illustrative temperatures $3000\mathrm{K}$ and $4500\mathrm{K}$ give $2.60268\times 10^5$ and $6.37793\times 10^6$ in the latter units. These are ideal predictions for stipulated temperatures.

The key change of variables is $\nu =c/\lambda$. Corresponding positive intervals carry the same power, so $B_{\lambda }d\lambda =B_{\nu }|d\nu |$ andEqual-width wavelength intervals have unequal widths in frequency; the factor $c/{\lambda }^2$ compensates for this. Merely replacing the horizontal-axis labels of $B_{\nu }$ by $c/\nu$ does not produce $B_{\lambda }$. This explains the frequency and wavelength peaks of Planck's law.

A short derivation uses the Boltzmann distribution for a single electromagnetic mode with thermal energies $nh\nu$, $n=0,1,\dots$. Put $q=e^{-h\nu /(k_{B}T)}$. The geometric series gives $Z={\sum }_{n\ge 0}{q}^n=(1-q{)}^{-1}$ and mean thermal energy $h\nu q/(1-q)$. The electromagnetic mode density in a box, including both independent polarization of an electromagnetic wave states of a photon, is $8\pi {\nu }^2/c^3$ per volume per unit frequency. Multiplication gives spectral energy density $u_{\nu }=8\pi h{\nu }^3/[c^3(e^{h\nu /(k_{B}T)}-1)]$; isotropy gives $B_{\nu }=c{u}_{\nu }/(4\pi )$. This counts thermal excitation energy, excluding zero-point energy. ^[ref]

Planck's law connects quantum mechanics to a measurable radiation curve. Its small-$h\nu /(k_{B}T)$ limit is the Rayleigh-Jeans law, $B_{\nu }\approx 2k_{B}T{\nu }^2/c^2$; its large-$h\nu /(k_{B}T)$ limit is the Wien approximation, $B_{\nu }\approx (2h{\nu }^3/c^2)e^{-h\nu /(k_{B}T)}$. The suppression at high frequency makes the total energy density finite. Integrating the curve gives the Stefan–Boltzmann law, while locating its peak gives Wien's displacement law. Thus one formula predicts both the spectral shape and the total emission.

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.

For radiation in vacuum with isotropy, energy is distributed equally per solid angle: write $u_{\Omega }=u/(4\pi )$ for energy density per unit solid angle. Let $\theta$ be the angle between a ray's path and the wall normal. Its normal velocity $c\cos \theta$ determines the arrival rate, while normal momentum per energy is $\cos \theta /c$. Their product is ${\cos }^2\theta$. For a perfectly reflecting cavity wall,The leading two is the normal momentum reversal; the incoming hemisphere contains half of the field's directional energy density. This is also the normal component of the bulk pressure tensor, $P_{nn}={\int }_{4\pi }{u}_{\Omega }{\cos }^2\theta d\Omega =u/3$. Isotropy makes all three diagonal components equal, while their sum is $u$ because the squared direction components sum to one. The factor one third therefore depends on three-dimensional angular geometry, not on the shape of the blackbody radiation spectrum.

A perfectly absorbing wall exposed to this incoming hemisphere receives only $u/6$ from absorption. If it is in thermal equilibrium with the isotropic radiation, its outgoing emission supplies another $u/6$ of recoil, recovering $u/3$. A cold absorber before equilibrium is a different boundary condition. Similarly, the one-way incident energy flux is $cu/4$, whereas the net energy flux across a plane in the undisturbed isotropic field is zero. Dividing $cu/4$ by $c$ misses the extra normal-momentum projection and does not give the cavity pressure.

For a stipulated isotropic radiation energy density $u=6\mu \mathrm{J}{\mathrm{m}}^{-3}$, the bulk pressure is $2\mu \mathrm{Pa}$ regardless of its spectral shape. In thermal equilibrium, blackbody radiation supplies $u=a_{\mathrm{rad}}{T}^4$ with radiation constant $a_{\mathrm{rad}}=4\sigma /c$, so $p_{\mathrm{rad}}=a_{\mathrm{rad}}{T}^4/3$; Stefan–Boltzmann law defines $\sigma$. The equilibrium photon gas derives this same equation of state from its free energy.^[ref]

In uniformly rotating stellar hydrostatic equilibrium, composition-uniform thermodynamic quantities are functions of the total potential. Radiative conductivity gives $\mathbf{F}=f(\varphi )\nabla \varphi$, with $f=-\chi dT/d\varphi$. On one surface, $f$ is constant, so the outward flux is $F_n=fg$. The Stefan–Boltzmann law therefore gives $T_e\propto g^{1/4}$. This is a surface-flux law: the material temperature constant on an interior equipotential is not the latitude-dependent emergent effective temperature.

For gas-pressure-supported stellar homology with $\kappa \propto {\rho }^{1/2}{T}^{-5/2}$ and specific nuclear heating $ϵ\propto \rho T^5$, the central scalings give $L_{\mathrm{nuc}}\propto M^7{R}^{-8}$ and radiative diffusion in a star gives $L_{\mathrm{rad}}\propto M^5{R}^{-1}$. Equating them yields the displayed laws. The Stefan–Boltzmann law gives $T_{\mathrm{eff}}\propto M^{29/28}$, hence $d\log L/d\log T_{\mathrm{eff}}=132/29$. Composition and dimensionless profiles are held fixed.

For a chemically uniform, ideal-gas-supported stellar homology sequence with nuclear energy generation $ϵ={ϵ}_0{X}^2\rho T^4$ and opacity $\kappa ={\kappa }_{0}Z(1+X)\rho T^{-4}$, radiative diffusion in a star fixes $LRZ(1+X)/({\mu }^8{M}^6)$ and nuclear heating fixes $X^2{\mu }^4{M}^6/(R^{7}L)$. Eliminating $L$ gives $R^6\propto X^{2}Z(1+X){\mu }^{-4}$, henceThe absence of $M$ in the radius law is a consequence of these particular power laws; it is not a universal relation for real stars. At fixed initial $X$ and $\mu$, the Stefan–Boltzmann law gives $T_e\propto Z^{-1/12}$ at fixed luminosity.

The equation of state of isotropic blackbody radiation is $p=\rho /3$. The barotropic hydrostatic first integral gives $\rho N^4=C$. The Stefan–Boltzmann law $\rho \propto T^4$ then yields the Tolman–Ehrenfest relation $TN=\mathrm{constant}$. Here $N=\sqrt{|K^a{K}_a|}$ for the stationary timelike Killing vector field; its normalization only changes the constants.

The Stefan-Boltzmann constant converts the fourth power of absolute temperature into blackbody radiant exitance:Here $k_B$ is the Boltzmann constant, $h$ the Planck constant and $c$ the speed of light. In the current International System of Units, these three constants have exact defining values, so $\sigma$ is exact as an expression in them and $\pi$. The displayed finite decimal is an approximation, not an exact decimal or a measured uncertainty. The 2022 recommended-constants table lists an ellipsis and an exact uncertainty for $\sigma$.^[ref]

A dimensional check uses $[k_B]=\mathrm{J}{\mathrm{K}}^{-1}$, $[h]=\mathrm{J}\mathrm{s}$ and $[c]=\mathrm{m}{\mathrm{s}}^{-1}$. Consequently $[k_B^4/(h^3{c}^2)]=\mathrm{J}{\mathrm{s}}^{-1}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}=\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$, exactly the units needed in the Stefan–Boltzmann law.

For an equilibrium photon gas, the energy density is $u=a_{\mathrm{rad}}{T}^4$ with radiation constant $a_{\mathrm{rad}}=4\sigma /c$. The radiation pressure is $p_{\mathrm{rad}}=u/3=4\sigma T^4/(3c)$. These describe the radiation filling an equilibrium cavity; $u$ has units $\mathrm{J}{\mathrm{m}}^{-3}$, whereas the outward radiant exitance through a small opening is $cu/4$. The factor $1/4$ comes from averaging propagation directions and their projected area over a radiation field with isotropy, not from reducing the opening's emissivity.

Approximate an isolated Schwarzschild black hole as emitting photon blackbody radiation from an area $16\pi G^2{M}^2/c^4$ at Hawking temperature $\hslash c^3/(8\pi G{k}_{B}M)$. The Stefan–Boltzmann law gives $P=\hslash c^6/(15360\pi G^2{M}^2)$. Integrating $c^{2}dM/dt=-P$ gives the displayed lifetime, about $2\times 10^{67}$ years for a solar mass. Greybody factors and additional emitted species modify the coefficient; the robust large-mass scaling is $M_0^3$, and the semiclassical calculation does not determine the final quantum-gravity endpoint.

// Target: general-relativity.bigb

The Stefan–Boltzmann law gives the total hemispherical radiant exitance of an ideal blackbody at absolute temperature $T$:Here $\sigma$ is the Stefan-Boltzmann constant. Total means integrated over all wavelengths; hemispherical means integrated over all outward directions with the projected area weighting. Thus $M_b$ is emitted power per actual emitting area, rather than directional radiance or spectral radiance. Absolute temperature must be in kelvin; raising a Celsius reading to the fourth power gives the wrong result. A uniform surface of area $A$ emits radiant flux $P_{\mathrm{emit}}=A{M}_b$. A real surface instead has $M=ϵ(T)\sigma T^4$, where $ϵ(T)$ is its total hemispherical emissivity. The older symbol $F$ in the formula above denotes this emitted radiant exitance.^[ref]

For an ideal blackbody, doubling absolute temperature multiplies radiant exitance by $16$. With the approximate Stefan-Boltzmann constant $5.670374419\times 10^{-8}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$, the radiant exitances at $300\mathrm{K}$ and $600\mathrm{K}$ are approximately $459.300$ and $7348.805\mathrm{W}{\mathrm{m}}^{-2}$. This steep response explains why thermal radiation becomes so consequential at high temperature. It also enables heat transfer through a vacuum, where thermal conduction through a gas and convection are absent.

A short derivation integrates Planck's law in its frequency form. For a blackbody, direction-independent spectral radiance is $B_{\nu }(T)=2h{\nu }^3/[c^2(e^{h\nu /(k_{B}T)}-1)]$, where $h$ is the Planck constant, $k_B$ the Boltzmann constant and $c$ the speed of light. The outward hemisphere contributes $\int \cos \theta d\Omega =\pi$, soExpanding $(e^x-1{)}^{-1}={\sum }_{n=1}^{\infty }{e}^{-nx}$, nonnegative terms allow termwise integration. Repeated integration by parts gives ${\int }_0^{\infty }{x}^3{e}^{-nx}dx=6/n^4$; the Riemann zeta function value $\zeta (4)={\pi }^4/90$ gives the remaining integral as ${\pi }^4/15$. Hence $M_b=\sigma T^4$ with the coefficient in Stefan-Boltzmann constant. The angular integral is essential: omitting it confuses radiance with radiant exitance.