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For a reciprocal antenna with matched polarization and load, the available antenna effective area is proportional to the directional antenna gain. For a lossless antenna, a thermal equilibrium argument fixes the constant: one polarization of Rayleigh-Jeans law radiance gives incident power $(k_{B}T/{\lambda }^2)\int A_{\mathrm{eff}}d\Omega d\nu$, and matched thermal electrical noise gives $k_{B}Td\nu$. Thus $\int A_{\mathrm{eff}}d\Omega ={\lambda }^2$. Since $\int Gd\Omega =4\pi$, the displayed relation follows. Reciprocal losses multiply both area and gain by the same efficiency.

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.

Consider an original idealized narrow-band model for one matched spatial mode and one polarization of waves. In the classical Rayleigh-Jeans law regime, an available thermal power can be written $P=k_{B}T\Delta \nu$, where $\Delta \nu$ is the effective bandwidth and $T$ the equivalent input noise temperature. Assume a linear detector with stable positive gain and absorb its bandwidth, Boltzmann constant and response factor into $g$. Then the output is $S=g(T+T_r)$, with input-referred receiver noise temperature $T_r$. These are model assumptions, not a reconstruction of the historical electronics.

Two references with known noise temperatures $T_h\ne T_c$ giveso the two-load radiometer calibration yieldsSubtracting the reference equations eliminates the unknown receiver term; substituting back determines it. An arbitrary stable output offset can likewise be eliminated from the interpolation for $T_s$, although separating that offset from $T_r$ then requires additional information.

For deliberately chosen inputs, set $T_h=300\mathrm{K}$, $T_c=4.2\mathrm{K}$, $T_r=40\mathrm{K}$ and $g=2mV/K$. These are equivalent noise temperatures; physical load thermometry must be converted consistently at the observing frequency. The reference outputs are $S_h=680\mathrm{mV}$ and $S_c=88.4\mathrm{mV}$. A sky-input output $S_s=90\mathrm{mV}$ gives $T_s=5.0\mathrm{K}$. If independent estimates assign $1.4\mathrm{K}$ to atmosphere, $0.5\mathrm{K}$ to antenna loss and $0.1\mathrm{K}$ to ground pickup in this lumped additive model, the residual is $3.0\mathrm{K}$.

The subtraction exposes what a noise budget must achieve. A $1\mathrm{K}$ error in the assigned atmospheric term becomes a $1\mathrm{K}$ error in the residual, even with exact detector calibration. Multiplicative attenuation requires a different correction: if an isothermal atmosphere has transmission $t$, its ideal contribution is $t{T}_{\mathrm{cosmic}}+(1-t)T_{\mathrm{atm}}$ in the same classical equivalent-temperature convention. Recovering $T_{\mathrm{cosmic}}$ then requires both subtracting atmospheric emission and dividing by $t$. The additive toy budget applies when its terms already refer to the same receiver input.

There is also a conditioning issue. If $S_h-S_c$ is small, a fixed output error strongly changes the inferred temperature. Widely separated reference temperatures improve the scale determination only if the receiver remains linear over that range; measuring the reference repeatedly tests stability rather than proving linearity. The experiment therefore motivates a model, a calibration and a set of independent checks, each with its own assumptions.

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.

The thermal Sunyaev-Zeldovich effect is inverse Compton scattering of cosmic microwave background photons by hot cluster electrons. For an optically thin, nonrelativistic thermal plasma,The Compton y parameter measures the electron pressure column. The distortion is a decrement below roughly 217 GHz, an increment above it, and has $\Delta T/T=-2y$ in the Rayleigh-Jeans law limit. Relativistic electron corrections modify the spectrum and null.^[ref]

Multifrequency maps separate this spectrum from primary Cosmic microwave background anisotropy, radio sources and dust; atmospheric subtraction, beams and calibration require control. The kinetic Sunyaev-Zeldovich effect has $\Delta T/T\simeq -\tau v_{\mathrm{los}}/c$ and a CMB-temperature-like spectrum, so it is both a contaminant and a peculiar-velocity probe. Planck demonstrated blind multifrequency cluster detection with external confirmation and a measured selection function.^[ref]

For a fixed physical pressure column, $y$ has no ordinary cosmological surface-brightness dimming: the distorted spectrum and CMB redshift together. The integrated Compton parameter $Y=\int yd\Omega$ still scales as $D_A^{-2}$ for fixed integrated physical pressure. SZ observations therefore find distant clusters and measure thermal pressure; integrated $Y$ is a useful mass proxy, requiring calibration and a selection model for cosmological counts. X-ray emission scales approximately as $\int n_e^2\Lambda (T_e)dl$, while SZ scales as $\int n_e{T}_{e}dl$. Combining them and a measured temperature constrains density, depth and, with angular geometry, distance. gas clumping factor, asphericity, temperature structure and nonthermal support limit that inference.

Choose the baseline vector from the first antenna to the second. Apart from a sign convention for receiver delay, the geometric delay is ${\tau }_g=\mathbf{b}\cdot \mathbf{s}/c$, and the phase difference is $\varphi =2\pi \nu {\tau }_g$. For equal received source powers $P_a$, adding the voltages and averaging over many wave periods givesUncorrelated receiver noise adds a mean level but not the source's coherent cross term. For an east-west baseline and one choice of hour-angle sign, ${\tau }_g=(b/c)\cos \delta \sin H(t)$. Earth rotation therefore sweeps the phase; locally the fringe frequency is $(b/\lambda )\cos \delta \cos H\dot{H}$. Tracking the antenna pointing keeps the source in the primary beam. Electronic fringe stopping is a separate operation: it can remove this phase evolution, making a source at the phase centre constant in the corrected complex visibility. The oscillating power above is the geometrical fringe before that correction.

For a small angular displacement $d\theta$ along the projected baseline, $d\varphi =2\pi b_{p}d\theta /\lambda$. One full fringe therefore spans $\Delta \theta \sim \lambda /b_p$. Here $b_p=|\mathbf{b}-(\mathbf{b}\cdot \mathbf{s})\mathbf{s}|$. This is the angular period of the measured Fourier component and the characteristic resolution scale when such long baselines are synthesized; precise beam width and image ambiguity also depend on baseline coverage and weighting. A single pair by itself does not give an unambiguous complete image.

Writing $x=h\nu /(k_{B}T)$, the expansion $e^x-1=x+O(x^2)$ immediately gives the Rayleigh-Jeans law,The factor two counts the two independent modes of polarization of waves. A small brightness-temperature perturbation thus has $\Delta I_{\nu }=2k_B{\nu }^2\Delta T_{\mathrm{RJ}}/c^2$ in this limit.

Hot intracluster electrons inverse-Compton scatter the microwave-background photons. The scattering optical depth is $d\tau =n_e{\sigma }_{T}dl$ and the characteristic thermal fractional energy transfer is of order $k_B{T}_e/(m_e{c}^2)$. The integrated distortion is therefore controlled byScattering conserves photon number while shifting photons toward higher frequencies, giving a low-frequency decrement and a high-frequency increment. Thus the printed estimate correctly gives a negative change of order $y{T}_{\mathrm{CMB}}$. Its numerical coefficient, however, is not the standard Rayleigh-Jeans coefficient. For nonrelativistic electrons and small $y$, the Kompaneets equation gives $\Delta n=y{x}^{-2}{\partial }_x(x^4{\partial }_x{n}_0)$. At small $x$, $n_0\simeq 1/x$, so $\Delta n\simeq -2y/x$ andEquivalently the thermodynamic-temperature spectral factor is $x\coth (x/2)-4\to -2$. The original PDF really prints the estimate with $-y$, not $-2y$; both numerical conventions are retained below. Relativistic corrections at the supplied $T_e$ are small compared with this factor-of-two difference.^[ref]

The uniform cube has a constant line-of-sight depth $L=6\times 10^{22}\mathrm{m}$. Using the supplied cross-section rather than replacing it with a more precise value givesAdopt $T_{\mathrm{CMB}}=2.72548\mathrm{K}$, explicitly an external measured constant, not a temperature supplied in the question. The values $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$ and $m_e\simeq 9.10938\times 10^{-31}\mathrm{kg}$ set the physical units. Hence the printed relation gives $\Delta T\simeq -22\mu \mathrm{K}$; the standard Rayleigh-Jeans result gives $\Delta T\simeq -44\mu \mathrm{K}$. Keeping the answer as $-8.14\times 10^{-6}{T}_{\mathrm{CMB}}$, or twice that, avoids needing a numerical background temperature.^[ref]

The face is a square, so its solid angle is $\Omega =(6\pi /10800{)}^2=3.046\times 10^{-6}\mathrm{sr}$, not the area of a circular six-arcminute aperture. At $15\mathrm{GHz}$,Thus $\Delta S_{\nu }\simeq -4.67\times 10^{-4}\mathrm{Jy}=-0.47\mathrm{mJy}$ using the printed estimate, or $-0.93\mathrm{mJy}$ using the standard factor two. The sign denotes missing brightness relative to the surrounding microwave background. At this frequency $h\nu /(k_B{T}_{\mathrm{CMB}})\simeq 0.264$; the exact nonrelativistic spectral intensity differs from the strict Rayleigh-Jeans approximation by about one percent. This precision does not affect the estimates.

For the array noise, the system temperature sets the receiver noise scale; let $A_e={\eta }_A\pi d^2/4$ be the effective area of each dish, and let ${\eta }_c$ describe correlation losses. No efficiencies or polarization of waves count are supplied, so first take ${\eta }_A={\eta }_c=1$ and a single independent polarization of waves. An unpolarized point source produces antenna temperature $T_a=S_{\nu }{A}_e/(2k_B)$, and the system-equivalent flux density is $\mathrm{SEFD}=2k_B{T}_{\mathrm{sys}}/A_e$. The interferometric radiometer equation gives one baseline an rms flux error $\mathrm{SEFD}/\sqrt{2\Delta \nu t}$. There are $N(N-1)/2=45$ cross-correlation baselines; averaging their independent thermal noise for a point source givesThe ideal single-dish effective area is $4\pi {\mathrm{m}}^2$, giving $\mathrm{SEFD}=6.59\times 10^3\mathrm{Jy}$ andCombining two independent parallel-hand polarization of waves measurements gives $6.3\mu \mathrm{Jy}$. Nonunit efficiencies increase these estimates by $1/({\eta }_A{\eta }_c)$. Here signal equal to noise means a weak-source signal-to-noise ratio of one after integration; it does not mean equating the source to the entire unaveraged receiver-noise power.^[ref]

At the band centre, $\lambda =0.01999\mathrm{m}$. Taking a representative maximum projected baseline of $20\mathrm{m}$ gives ${\theta }_{\mathrm{res}}\sim 3.4$ arcminutes. A corner-to-corner baseline of $20\sqrt{2}\mathrm{m}$ would give about $2.4$ arcminutes before projection and weighting. The exact antenna locations and source declination are not supplied, so neither a unique synthesized beam shape nor an exact FWHM is determined. The individual-dish primary-beam scale, roughly $\lambda /4\mathrm{m}\sim 17$ arcminutes, is the field of view, not the synthesized resolution.

To extract the cluster signal, retain frequency-resolved complex visibilities and fit an extended SZ template together with contaminating compact sources. For the ideal uniform square centred at the phase centre and with a nearly constant primary beam across it, the Fourier transform isThis is why the integrated decrement must not be treated as a point-source flux on every baseline. The first axial visibility null occurs at $b=\lambda /\Theta \simeq 11.5\mathrm{m}$. Short baselines carry most of the large-scale decrement; the longest baselines are useful for separating unresolved positive radio sources. natural visibility weighting or a suitable taper gives a clearer extended-source image than emphasizing the longest baselines, while a visibility-domain matched filter uses all samples with weights proportional to template amplitude divided by noise variance. Its amplitude estimate is $\hat{a}=\mathrm{Re}(T^{†}{C}^{-1}d)/(T^{†}{C}^{-1}T)$ for a real template amplitude. A joint fit propagates radio-source uncertainty instead of subtracting exact, noise-free source models.

Earth-rotation and multifrequency synthesis improve Fourier coverage, but each channel must use its own baseline in wavelengths, primary beam and SZ spectrum. Use the longer baselines and spectra to identify variable point sources, flag bad data, and test the decrement with time, antenna and frequency splits. Primary-beam and absolute-flux calibration must be included. Large-scale microwave-background fluctuations and residual ground or atmospheric signals can remain even below the thermal-noise estimate. The narrow low-frequency band alone gives limited separation of SZ and primary microwave-background temperature structure. There is no zero-spacing interferometric datum: integrated flux is model dependent unless sufficiently short baselines or additional total-power information are available. Exact cluster detection significance cannot be obtained from the point-source noise alone without the actual projected baseline distribution and weighting.

An electron accelerated in an ion's Coulomb field emits bremsstrahlung. Both electron states are unbound, giving free-free radiation rather than a discrete bound-state line. Averaging over a Maxwellian electron distribution gives an optically thin spectral emissivity with dependenceHere $C_{\mathrm{ff}}$ depends on the emissivity convention, and the Gaunt factor supplies the quantum correction. No numerical emissivity normalization is needed for these diagnostics. For a mixture, sum $Z_i^2{n}_i$ over ion species. The bolometric emissivity is proportional to $n_e\sum Z_i^2{n}_i{T}^{1/2}$ with a frequency-averaged Gaunt factor. Thermal electron speeds control the spectrum while the encounter rate supplies its density-squared dependence.

temperature can be measured from the high-energy exponential cutoff at photon energy of order $k_{B}T$, provided the spectrum spans that turnover and foreground absorption and nonthermal contamination are modelled. At radio frequencies $h\nu \ll k_{B}T$, the exponential is nearly one, so the optically thin continuum is a poor thermometer by slope alone. Kirchhoff's law, ${\alpha }_{\nu }=j_{\nu }/B_{\nu }$, and the Rayleigh-Jeans law give ${\alpha }_{\nu }\propto n_e{n}_i{T}^{-3/2}{\nu }^{-2}{\bar{g}}_{\mathrm{ff}}$ for an isothermal thermal plasma. The transfer solution isin the radio limit. A resolved optically thick region therefore measures its electron temperature through brightness temperature, if it fills the beam. The turnover between thick and thin emission constrains the line-of-sight emission measure $\mathrm{EM}=\int n_e^{2}dl$ for a hydrogen-dominated fully ionized plasma. In the thin regime $T_b\simeq T{\tau }_{\nu }$, rather than $T_b=T$. The slowly varying Gaunt factor gives a nearly flat optically thin radio flux-density spectrum, in contrast to the ${\nu }^2$ thick limit at fixed projected area.^[ref]

To obtain density, one also needs geometry. An angular extent and independent distance give a transverse size; an assumed shape provides a depth or volume. For approximately uniform emitting gas with astrophysical volume filling factor $f_V$, $L_{\mathrm{ff}}\propto n_e^2{f}_{V}V{T}^{1/2}$, so $n_e\propto [L_{\mathrm{ff}}/(f_{V}V{T}^{1/2}){]}^{1/2}$. Alternatively a measured column emission measure and assumed path length give an rms density. Clumping raises the mean squared density relative to the square of the mean: the inferred value is not automatically the mean gas density. Free-free brightness alone does not uniquely determine temperature, density and physical size. Resolved imaging, a distance, optical-depth information or an independent column-density diagnostic is needed to remove those degeneracies. Cluster SZ data are complementary because their leading signal depends on $\int n_e{T}_{e}dl$, whereas free-free emission depends on density squared. Metal lines, unresolved temperature structure, foreground absorption, synchrotron emission and background subtraction must be controlled before attributing the continuum to a single thermal component.

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.

antenna quantities. radiation resistance represents emitted power by $P_{\mathrm{rad}}=I_{\mathrm{rms}}^2{R}_{\mathrm{rad}}$ at the feed. It is not Joule heating in the antenna; a separate loss resistance accounts for that. antenna gain is the directional far-field intensity relative to an isotropic radiator supplied with the same input power:It includes radiation efficiency and equals antenna directivity only for a lossless antenna. antenna effective area is defined by the power available to a matched load, $P_{\mathrm{av}}=A_{\mathrm{eff}}{S}_{\mathrm{inc}}$, for incidence from a specified direction and matched polarization. It need not equal a literal geometric collecting area.

Gain–area relation. To derive the antenna gain-effective-area relation, take a passive reciprocal antenna. Its electromagnetic reciprocity makes the receive pattern proportional to the transmit pattern; write $A_{\mathrm{eff}}(\Omega )=KG(\Omega )$. The normalization follows from thermal equilibrium. In the Rayleigh-Jeans law limit, blackbody radiation radiance per unit frequency per unit solid angle is $2k_{B}T/{\lambda }^2$ for two polarizations. A single antenna polarization couples to one half, so the available incident noise power isThe open-circuit Johnson voltage spectrum of the antenna resistance is $4k_{B}T{R}_{\mathrm{rad}}d\nu$. A matched load receives half the open-circuit voltage, giving available power $k_{B}Td\nu$. In a lossless antenna at thermal equilibrium these powers agree; hence $\int A_{\mathrm{eff}}d\Omega ={\lambda }^2$. Since $\int Gd\Omega =4\pi$,The same efficiency factor multiplies receiving area and gain if reciprocal dissipative losses are included consistently. Impedance mismatch or polarization mismatch reduces delivered power further. The factor of two for thermal polarization modes and the factor of four between open-circuit voltage variance and matched-load power are separate effects.

Normalize the radar beam. Interpret the azimuth as signed about the beam centre at $\varphi =0$, so both sides of the narrow lobe are included. Put $u=\theta -\pi /2$. To leading order in the small angular width, $\sin \theta \simeq 1$ and the angular integrals can be extended over the real line:The Gaussian antenna beam solid angle gives the peak on the positive $x$-axis consequently hasHere $N$ is power per solid angle, equal to $r^2{S}_r$ in the far field. The leading neglected solid-angle correction is relative order ${\alpha }^2$.

Return using the supplied electric-dipole response. The electric-dipole radar cross section of a small conducting sphere determines this return. At target range $D$, the incident intensity is $S_t=C/D^2$. Write its root mean square electric field as $E_t$, with $S_t={ϵ}_{0}c{E}_t^2$. The induced electric dipole is $p_{\mathrm{rms}}=4\pi {ϵ}_0{a}^3{E}_t$. A dipole's angular radiation pattern, normalized by $\int {\sin }^2\chi d\Omega =8\pi /3$, gives $d{P}_{\mathrm{sc}}/d\Omega =(3P_{\mathrm{sc}}/8\pi ){\sin }^2\chi$. The backward direction is perpendicular to the induced dipole, so the supplied total dipole power givesThe received intensity is this angular power divided by $D^2$. Using the monostatic radar equation, or directly multiplying by the receive area, givesThe receiver threshold is a voltage across the matched load, so the detectable power is $V_m^2/R$, without another matching factor:This is the electric-dipole result following from the response data provided. It assumes $ka\ll 1$, far-field propagation on each leg, alignment at the beam maximum and no attenuation or additional receiver losses.

Perfect-conductor qualification. perfect-conductor sphere dipole interference arises because the sphere also excludes the oscillating magnetic field. Solving the magnetic boundary problem gives $\mathbf{m}=-2\pi a^3{\mathbf{H}}_t$ in addition to the electric dipole. The magnetic polarizability of a perfectly conducting sphere follows from its induced surface currents.^[ref] In the backward direction the magnetic contribution has half the electric scattering amplitude and the same sign, so the returned intensity is $9/4$ of the electric-only value. Equivalently, the monostatic radar cross section is ${\sigma }_{\mathrm{back}}=9\pi k^4{a}^6$, rather than $4\pi k^4{a}^6$. Including both leading dipoles therefore givesBoth results have been distinguished because the supplied auxiliary data describe only the electric dipole, while the literal perfect-conductor boundary conditions also generate the magnetic one. The backward interference is a leading-order effect, not a higher-order correction in $ka$.^[ref]