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For a uniform spherical cluster of depth $\ell \propto \theta d_A$, thermal bremsstrahlung gives received bolometric flux $S_X\propto n_e^2{T}_e^{1/2}{\ell }^3/d_L^2$, while the Rayleigh-Jeans law limit of the thermal Sunyaev-Zeldovich effect gives $|\Delta T|\propto n_e{T}_e\ell$. Eliminating $n_e$ and using $d_L=(1+z{)}^2{d}_A$ yields the displayed angular diameter distance. A model for the density profile fixes its numerical coefficient. Clumping, nonsphericity and temperature structure can bias this distance because the two observables weight the gas differently.

For blackbodies with $T_p<T_s$, the contrast $r_{\lambda }=(R_p/R_s{)}^2(e^{x_s}-1)/(e^{x_p}-1)$ increases with wavelength, where $x_i=hc/(\lambda k_B{T}_i)$. Indeed $d\log r/d\log \lambda =G(x_p)-G(x_s)>0$, with $G(x)=x/(1-e^{-x})$ increasing for $x>0$ because $1-(1+x)e^{-x}>0$. The contrast approaches $(R_p/R_s{)}^2{T}_p/T_s$ in the Rayleigh-Jeans law limit. This behavior differs from the peaked absolute blackbody flux.

Write $n=n_0+y\mathcal{L}{n}_0+O(y^2)$, with $n_0=(e^x-1{)}^{-1}$ and $\mathcal{L}=x^{-2}{\partial }_x(x^4{\partial }_x)$. Differentiating the Planck photon distribution givesDividing by $n_0$ proves the first-order thermal Sunyaev-Zeldovich effect:In the Rayleigh-Jeans law limit, $x{e}^x/(e^x-1)\to 1$ and $x\coth (x/2)\to 2$, so $\Delta n/n_0\to -2y$. In the Wien approximation, the prefactor approaches $x$ and the bracket approaches $x-4$, so $\Delta n/n_0\sim yx(x-4)\sim y{x}^2$. The last limit is an asymptotic description of the first-order coefficient: a perturbative prediction at large $x$ still requires $y{x}^2\ll 1$.

Hot electrons transfer energy to the cosmic microwave background, removing low-frequency photons and enhancing the high-frequency tail. The first-order distortion crosses zero where $x\coth (x/2)=4$, at $x\simeq 3.830$; for today's CMB this is near $217\mathrm{GHz}$. Photon conservation follows directly from the boundary termIn contrast, integration by parts gives $\int x^3\Delta ndx=4y\int x^3{n}_{0}dx$, so radiation gains a fraction $4y$ of its original energy. The thermal Sunyaev-Zeldovich effect is thus a spectral distortion, not simply a new blackbody temperature.

A galaxy cluster with bulk peculiar velocity along the line of sight additionally produces the kinetic Sunyaev-Zeldovich effect. Define $v_{\parallel }>0$ for recession and $\tau =\int n_e{\sigma }_{T}d\ell$. To first order in velocity and optical depth, $\Delta T/T=-\tau v_{\parallel }/c$: receding gas gives a decrement and approaching gas an increment. Its occupation number perturbation is $(\Delta T/T)x{e}^x/(e^x-1{)}^2$, the spectrum of a small blackbody temperature change. It remains nonzero at the thermal null and therefore complicates measurements near that frequency.

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

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

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

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

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

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

For the Rayleigh-Jeans spectrum of a finite blackbody disk, the Rayleigh-Jeans law replaces $(e^{h\nu /(kT)}-1{)}^{-1}$ by $kT/(h\nu )$. Here $h$ is the Planck constant and $k$ the Boltzmann constant. The multitemperature blackbody disk therefore hasFor example, setting $R=r_{\mathrm{out}}/r_{\ast }$ makes its frequency-independent coefficient proportional to the finite dimensionless integralAlthough the exact effective temperature vanishes at the inner edge, the very narrow cold rim where the Rayleigh-Jeans law fails makes a negligible contribution in this limit. More formally, after dividing the integrand by ${\nu }^2$, the inequality $e^x-1\ge x$ bounds it by $(k/h)rT$, so dominated convergence justifies the result even at that edge.

At large radius, $T\propto r^{-3/4}$, and the contribution per logarithmic interval is $r^{2}T\propto r^{5/4}$. The outer disk dominates the low-frequency emission because its much greater area outweighs its lower effective temperature.

For intermediate frequencies use the allowed power-law approximation to the effective temperature and introduceThensoThe lower limit is much smaller than one and the upper limit much larger than one. Extending them to zero and infinity leaves a constant: near zero the integrand behaves as $x^{2/3}$, and at infinity it decays exponentially. Hence the intermediate spectrum isThis is the one-third spectrum of a multitemperature disk. Much of the emission comes from radii where $kT(r)$ is of order $h\nu$, moving inward as the frequency rises. With the exact inner-edge profile, a broad intermediate interval also requires frequency well below $k{T}_{\max }/h$; the supplied approximation captures its slope away from the hottest annuli.

Assume $R_p/R_{\star }=0.10$, $T_{\star }=5800\mathrm{K}$, uniform disk-averaged thermal emission, a nearly blackbody stellar continuum, negligible reflected light at $20\mu \mathrm{m}$, and a flat response over the narrow band. If the eclipse depth is normalized to the out-of-eclipse star-plus-planet flux, the planet/star flux ratio is $r=0.005/(1-0.005)=0.005025$; using $r=0.005$ changes the estimate by less than one percent. The exoplanet secondary eclipse then givesUsing the Planck law at $\lambda =20\mu \mathrm{m}$ gives $T_1\simeq 3.1\times 10^3\mathrm{K}$, or about $3000\mathrm{K}$ at the precision justified by the radius assumption. The Rayleigh-Jeans law instead gives $T_1\simeq 0.5T_{\star }\simeq 2900\mathrm{K}$, a useful rough check. This is a brightness temperature, not the bolometric planetary equilibrium temperature. With zero Bond albedo and global redistribution, the latter would be only about $1250\mathrm{K}$ at $0.05\mathrm{AU}$; a thermal window can sample hotter, deeper gas.

A continuum photon normally emerges from an optical depth of order unity. In hydrostatic equilibrium,The emission-pressure degeneracy of a brightness temperature means that pressure cannot be uniquely recovered from the measured brightness temperature without $g$ and an opacity model. For an explicit nominal estimate, take $g=20\mathrm{m}{\mathrm{s}}^{-2}$ and an effective continuum mass absorption coefficient ${\kappa }_{20}=2\times 10^{-4}{\mathrm{m}}^2\mathrm{k}{\mathrm{g}}^{-1}$. Then $P_{\tau \sim 1}\sim g/{\kappa }_{20}=10^5\mathrm{Pa}$, or approximately one bar. In a hydrogen-rich atmosphere, collision-induced absorption by transient ${\mathrm{H}}_2$ pairs supplies continuum opacity even without ordinary molecular lines; its density dependence makes the constant-opacity calculation an illustrative estimate. With a normal deep atmospheric pressure-temperature profile, temperature increases at larger pressures, eventually following an interior adiabatic temperature gradient.

Two mechanisms for $T_1\ne T_2$ are wavelength-dependent sampling of a vertically varying temperature, and wavelength-dependent weighting of a horizontally nonuniform dayside. In the first, carbon monoxide-rich opacity near $4.5\mu \mathrm{m}$ can sample cooler gas around $0.1$ bar while the $20\mu \mathrm{m}$ continuum samples hotter gas around one bar. In the second, hot and cool areas contribute different proportions of the total flux in different bands because the Planck function depends nonlinearly on temperature. The vertical mechanism alone produces the illustrated pair of temperatures; a single isothermal, horizontally uniform blackbody would have the same brightness temperature in both bands.

If the J-band observation is thermal emission, its low-opacity window normally samples at least as deeply as the continuum at $20\mu \mathrm{m}$. On the illustrated non-inverted profile, $T_3\gtrsim T_1>T_2$. The ordering is conditional: molecular-line absence alone does not specify the relative continuum opacities, and an inversion or reflected-light contamination can change the inference. The exoplanet emission spectrum measures weighted contribution regions rather than three exact thermometers at known pressures.

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

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

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

At exoplanet secondary eclipse, the full-phase planet-star flux ratio is the sum of reflected and thermal light. Approximating both bodies as unresolved blackbodies and taking wavelength-independent geometric albedo,The first term is a flat reflected-light level under the stated constant-albedo assumption. At short wavelength the cool planet lies in the Wien limit, so thermal emission is exponentially suppressed and reflection dominates. At long wavelength both spectra enter the Rayleigh-Jeans law, givingThe sketch therefore starts on the reflected plateau, rises where planetary blackbody radiation becomes important, and asymptotically approaches the long-wavelength plateau. This neglects spectral albedo features, phase dependence, stellar lines, and a nonisothermal planetary photosphere.

A homogeneous interpretation assigns the planet a brightness temperature $T_b$ satisfyingAt a fixed wavelength let $x=hc/(\lambda k_B)$ andInverting the Planck law givesAt $\lambda =20\mu \mathrm{m}$, $x=719.4\mathrm{K}$. With ${\theta }_T=\pi /4$, $T_1=1500\mathrm{K}$, and $T_2=1000\mathrm{K}$, one obtainsThe Rayleigh-Jeans law would give the nearly identical area-weighted estimate $1250\mathrm{K}$.

At secondary eclipse the removed monochromatic planetary flux is $\pi B_{\lambda }(T_p)R_p^2/d^2$, while the stellar flux is $\pi B_{\lambda }(T_s)R_s^2/d^2$. The thermal exoplanet secondary eclipse depth is thereforeA hot Jupiter has tiny visible thermal contrast, rapidly improving contrast toward the infrared, and molecular absorption or emission features superposed on its continuum. Reflected light may add a visible component. In the Rayleigh-Jeans law limit, $B_{\lambda }(T)\propto T/{\lambda }^4$, soa wavelength-independent asymptote for ideal blackbodies.

The peak near $0.5\mu \mathrm{m}$ can be reflected starlight carrying the stellar spectral shape, while the $1\mu \mathrm{m}$ peak can be thermal emission from a very hot, young, or strongly irradiated planet. Wien displacement law then suggests $T_s\simeq 5800\mathrm{K}$ and $T_p\simeq 2900\mathrm{K}$, so $T_p/T_s\simeq 1/2$.

Assume both objects are in the far-infrared Rayleigh-Jeans law regime. Fromone obtains $R_p/R_s\simeq \sqrt{2\times 10^{-3}}\simeq 0.045$. With $R_s=0.5R_{\odot }$ and $R_J\simeq 0.10R_{\odot }$,of order a Neptune radius. This estimate is sensitive to the peak interpretation and blackbody assumptions.

If the system transits, secondary-eclipse emission spectroscopy is especially effective because it separates planetary light from starlight and directly measures the favorable $10^{-3}$ infrared contrast. For a sufficiently wide orbit, direct-imaging spectroscopy is preferable because the hot planet is self-luminous and can be spatially separated from its star.

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.

For blackbody star and planet observed at one common wavelength sufficiently long for the Rayleigh-Jeans law to hold for both,This follows by substituting $B_{\lambda }=2c{k}_{B}T/{\lambda }^4$ in the thermal eclipse depth. A radiative flux ratio alone does not determine the radius without a planetary brightness temperature and a stellar temperature. Spectral peaks interpreted with Wien's displacement law must refer to $F_{\lambda }$, not $F_{\nu }$ or $\lambda F_{\lambda }$.

For a finite multitemperature blackbody disk, the low-frequency Rayleigh-Jeans law gives $F_{\nu }\propto {\nu }^2\int rT(r)dr$. If $T\propto r^{-3/4}$ away from the inner edge, outer annuli dominate because the weight per logarithmic radial interval grows as $r^{5/4}$. A measure-zero zero-temperature inner edge does not invalidate the leading law when $rT(r)$ is integrable.

For a Planck photon distribution, the hot-electron Kompaneets equation gives $\Delta n=yx{e}^x(e^x-1{)}^{-2}[x\coth (x/2)-4]$, where $x=h\nu /(k_B{T}_{\gamma })$. The equivalent thermodynamic temperature perturbation has the displayed form, while the Rayleigh-Jeans law gives $\Delta T/T=-2y$. The distortion changes sign at $x\simeq 3.830$. This first-order expression assumes a small Compton y parameter and nonrelativistic thermal Electrons.