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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.

At a fixed temperature, the frequency maximizing $B_{\nu }$ and the wavelength maximizing $B_{\lambda }$ in Planck's law are different ways of describing the same black-body radiation. They are maxima of different densities. The inverse relation between frequency and wavelength does not preserve those maxima because its interval-conversion factor varies with wavelength.

For a short extremum calculation, substitute $x=h\nu /(k_{B}T)$ into $B_{\nu }$, leaving a positive constant times $x^3/(e^x-1)$. Substitute $x=hc/(\lambda k_{B}T)$ into $B_{\lambda }$, leaving a positive constant times $x^5/(e^x-1)$. Setting each derivative to zero givesSelect the positive roots: $x_{\nu }=2.821439372\dots$ and $x_{\lambda }=4.965114232\dots$. Although $x=0$ solves the rearranged equations, it is an endpoint, not either peak. For $m=3$ or $5$, the function $m(1-e^{-x})-x$ initially increases, its derivative $m{e}^{-x}-1$ crosses zero once, and it eventually becomes negative. Hence it has exactly one positive root. Each density tends to zero at both ends, so this root gives its unique maximum. The bisection method brackets these roots in $(1,3)$ and $(1,5)$.

Consequently Wien's displacement law givesThese constants agree with NIST's tabulation. ^[ref] The wavelength peaks at $3000$, $4500$, and $6000\mathrm{K}$ are $965.924$, $643.949$, and $482.962\mathrm{nm}$. Doubling temperature halves this peak wavelength.

At $6000\mathrm{K}$, the frequency peak is ${\nu }_{\max }\approx 352.736\mathrm{THz}$, whose converted wavelength is $c/{\nu }_{\max }\approx 0.849907\mu \mathrm{m}$. The $B_{\lambda }$ peak is instead $0.482962\mu \mathrm{m}$, corresponding to about $620.737\mathrm{THz}$. These numbers are compatible: equal-width frequency bins and equal-width wavelength bins partition the same radiation differently. Specifying which density was maximized is essential when using a peak to infer temperature.

Infrared is electromagnetic radiation with wavelengths longer than the red end of ordinarily visible light. A useful named convention is the CIE range $780\mathrm{nm}$ to $1\mathrm{mm}$; the approximate visible range $400$–$700\mathrm{nm}$ used in some teaching texts puts the boundary nearer $700\mathrm{nm}$. ^[ref]. Neither number is a sharp physical cutoff of human vision. The CIE divisions IR-A, IR-B and IR-C are respectively $0.78$–$1.4\mu \mathrm{m}$, $1.4$–$3\mu \mathrm{m}$ and $3$–$1000\mu \mathrm{m}$; the boundaries of near-, middle- and far-infrared vary between applications. ^[ref].

Use vacuum wavelength $\lambda$ when comparing these ranges. With frequency $\nu$, speed of light $c$, Planck constant $h$ and photon energy $E$,Thus the CIE infrared interval corresponds approximately to $3.00\times 10^{11}$–$3.84\times 10^{14}\mathrm{Hz}$ and $1.99\times 10^{-22}$–$2.55\times 10^{-19}\mathrm{J}$ per photon, with the smaller energy at the longer wavelength. A $940\mathrm{nm}$ infrared photon has $E\simeq 2.11\times 10^{-19}\mathrm{J}$, whereas a $254\mathrm{nm}$ ultraviolet photon has $E\simeq 7.82\times 10^{-19}\mathrm{J}$. These are calculations from $hc/\lambda$, not measurements of a particular source. A television remote is a familiar approximately $940\mathrm{nm}$ source; its information is carried by variations in the emitted radiant flux. ^[ref].

Detection and visual perception answer different questions. A photodetector can respond to infrared even when an observer perceives no light; a thermometer detects an increase in temperature after photon absorption. Conversely, a small readout can reflect a small spectral responsivity, rather than absence of incident electromagnetic radiation. Images made from infrared readouts use display colors chosen by the imaging system; those colors are not the colors of the incoming infrared.

The important microscopic mechanisms also differ. In infrared spectroscopy, photon absorption often increases the energy of a normal mode in which chemical bonds stretch or bend. The corresponding optical spectrum helps distinguish molecules by their motions. For the usual electric dipole mechanism, the motion must change the electric dipole moment; matching the frequency alone is not sufficient. ^[ref]. Electronic processes can also emit or absorb infrared, so it is not exclusively radiation from molecular motion. ^[ref]; ^[ref].

Infrared is especially useful for observing thermal radiation from objects near everyday temperatures. For an ideal $300\mathrm{K}$ blackbody, Wien's displacement law gives a wavelength peak near $9.66\mu \mathrm{m}$. A thermal camera's readout nevertheless depends on emissivity, reflected surroundings and transmission along the path, as well as temperature. ^[ref]. A source emitting infrared need not be hot: an infrared remote is useful precisely because a photodetector can distinguish its signal from a background. Visible light and ultraviolet can also cause heating when absorbed; “infrared” names a wavelength range, not a unique form of heat.

Herschel's 1800 discovery made the distinction between sight and detection experimentally concrete. His report describes a optical prism spreading sunlight across a white, ruled receiving surface, with three thermometers at different positions near the end of the visible optical spectrum. He observed heating beyond the red edge where no visible color appeared. The result established that sunlight contained invisible radiation less strongly refracted by that optical prism than the visible red; it did not measure a photon energy or establish today's infrared boundaries. ^[ref]; ^[ref].

A documented modern version by Caltech/IPAC uses a glass optical prism, white paper in a shaded box and three blackened liquid-filled thermometers. Initial readings in shade provide each thermometer's baseline; subsequent readings compare blue, yellow and just beyond red after the bulbs have had time to warm. The measured quantity is the rise in temperature, not a “temperature of a color”. The lesson reports the expected largest rise beyond visible red and specifically discusses matching the bulb coatings and thermometer calibration. Clouds and changes in the projected optical spectrum affect the comparison. ^[ref].

For a stronger modern interpretation, an additional shaded thermometer would monitor ambient drift, exchanging bulbs between positions would test differences between thermometers, and blocking the incident beam would distinguish beam-dependent heating from warmed supports or air. These are suggested controls, not claims about Herschel's historical procedure. Thermal conduction through the mounting and convection can change the readout. A broad bulb also samples a range of wavelengths; the optical prism's transmission and optical dispersion, the bulb's absorption and heat loss all affect the position of the largest rise. In particular, the result does not show that an infrared photon carries more energy than a visible photon.

A simple analytical illustration makes those limitations explicit. Let $\theta =T-T_{\mathrm{amb}}$ be the bulb's excess temperature, $C$ its heat capacity, $G$ its effective heat-loss coefficient in $\mathrm{W}{\mathrm{K}}^{-1}$, and $P_{\mathrm{abs}}$ the absorbed power. Assume one uniform bulb temperature, constant ambient temperature, constant $C,G$, and losses linear in $\theta$. Thenfor a beam switched on at $t=0$ with $\theta (0)=0$. After switching the beam off at $t=t_s$, $\theta (t)=\theta (t_s)e^{-(t-t_s)/\tau }$. The exponential function follows by solving this linear differential equation. The steady excess temperature measures $P_{\mathrm{abs}}/G$; it is not a direct measurement of the optical spectrum. With illustrative $C=0.020\mathrm{J}{\mathrm{K}}^{-1}$ and $G=0.0010\mathrm{W}{\mathrm{K}}^{-1}$, $\tau =20\mathrm{s}$. Absorbed powers of $1.0$ and $0.5\mathrm{mW}$ give steady rises of $1.0$ and $0.5\mathrm{K}$, regardless of which wavelengths supplied that absorbed power.

The upper schematic represents the logic of the optical prism experiment; its light ray positions are not a calibrated wavelength scale. The lower curves use the illustrative heat-balance model, with the beam blocked at $60\mathrm{s}$; they are not historical or modern measured data. The beauty of the experiment is that a simple change of detector extends observation beyond what the eye can perceive.

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.

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 $\lambda F_{\lambda }$.