Multi-Wavelength Observation and Atmospheric Windows

The Universe broadcasts on every channel at once. Cold molecular clouds hum in the millimetre, stars glow in the optical and near-infrared, stellar coronae and shocked supernova gas blaze in the X-ray, and accreting black holes fling photons into the gamma-ray, where a single quantum can outweigh a proton at rest. Between the longest radio wave and the hardest gamma ray lies more than twenty powers of ten in photon energy — and no astrophysical object tells its whole story in one decade of it. A source's temperature, through Wien's law, all but dictates the band in which it shines brightest, so to read its physics you must read it across bands. The obstacle is the air. Earth's atmosphere is a selective filter, transparent in only two broad slots — the optical window from roughly 300 nm to 1100 nm and the radio window from about a centimetre to tens of metres — plus partial infrared gaps between crushing water-vapour bands. Everywhere else the sky is walled off: ozone eats the ultraviolet, water and carbon dioxide devour the infrared, the air column photoelectrically swallows X-rays and pair-produces on gamma rays, and the ionosphere mirrors the longest radio waves back into space. Each closed band was pried open only by climbing above the layer responsible — a dry mountaintop, a balloon, an aircraft, and finally orbit. This article is the observational keystone of the head: it takes the flux and photometric machinery of the preceding articles and asks where in the spectrum can we stand, and what must we build to see the rest. The transmission law $I_\nu = I_\nu^{(0)}e^{-\tau_\nu}$ , the airmass scaling of ground-based photometry, the ionospheric plasma cutoff, and the diffraction limit that forces radio dishes to gigantic size are all developed here.

Key takeaways

Earth's atmosphere transmits only through the optical window ( $\sim$ 300–1100 nm), the radio window ( $\sim$ 1 cm to tens of m), and a few partial infrared gaps; everything else reaches the ground attenuated to nothing.
The blocking is molecule-specific: ozone ( $\mathrm{O_3}$ ) absorbs the ultraviolet, water vapour ( $\mathrm{H_2O}$ ) and carbon dioxide ( $\mathrm{CO_2}$ ) absorb the infrared, air atoms photoelectrically absorb X-rays and pair-produce on gamma rays, and the ionosphere reflects radio waves below its plasma frequency.
Transmission follows $I_\nu = I_\nu^{(0)}e^{-\tau_\nu}$ ; the optical depth scales with airmass, $\tau_\nu(z)=\tau_\nu(0)\sec z$ , giving the Bouguer line $m(X)=m_0+kX$ with $k=1.086\,\tau_0$ .
Wien's law $\lambda_{\max}T=2.898\times10^{-3}\ \mathrm{m\,K}$ ties temperature to band, so where a source is best observed and whether the atmosphere permits it are two sides of one number.
Each band demands its own technology — grazing-incidence X-ray mirrors, pair-tracking gamma-ray detectors, cryogenic bolometers, coherent radio amplifiers — because photon energy spans twenty orders of magnitude.
Key idea: Earth's atmosphere is transparent only in narrow optical and radio windows (plus partial IR bands); the rest of the electromagnetic spectrum can only be observed from space, and each band requires its own detector technology.

Where this lives today

Multi-wavelength astronomy is now the default, and its frontier is multi-messenger science, where opening every window on the same event is the whole game. When the gravitational-wave event GW170817 — a neutron-star merger — was detected in 2017, telescopes across the spectrum swung to it within hours: a short gamma-ray burst caught by Fermi from orbit, an optical–infrared kilonova tracked from the ground through the atmospheric window, ultraviolet from Swift above the ozone, and X-rays and radio emerging over days from Chandra and the VLA. No single band could have shown that a kilonova forges heavy elements or that the merger launched a relativistic jet; the physics lived in the joint spectral energy distribution assembled across windows demanding variously orbit, dry sites, and ground arrays. The same logic drives today's flagships — the Rubin Observatory scanning the optical nightly, JWST reaching the mid-infrared from L2, ALMA exploiting the driest sub-millimetre gaps at 5000 m in the Atacama. The atmospheric-window map is not history but the daily constraint deciding which photons of a transient reach a mountaintop and which need a rocket.

The physical picture

Picture the atmosphere as a coloured, semi-opaque curtain across the sky. It is clear in the visible — no accident, since eyes evolved to use the wavelengths the air transmits and where the Sun peaks — and clear again in the radio, which is why radio astronomy was born on the ground. Between and beyond those slots the curtain is a wall built of different materials at different heights: ozone in the stratosphere for the ultraviolet, water vapour and carbon dioxide lower down for the infrared, the whole air column for X-rays and gamma rays, and the ionised upper atmosphere mirroring the longest radio waves. To open a closed band you must rise above the layer that closes it. The first figure is the map every observational astronomer carries in their head.

Figure

Atmospheric opacity across the electromagnetic spectrum: radiation reaches the ground only through the optical window 300 nm to 1100 nm and the radio window about 1 cm to tens of metres — gamma, X-ray, ultraviolet, most infrared and sub-millimetre are absorbed and must be observed from space or high altitude

The wall is not uniform, and that non-uniformity is the whole subject. A source's temperature fixes its peak wavelength through Wien's law, and that peak may fall in an open window or deep behind the wall. Cold interstellar dust at tens of kelvin peaks in the far-infrared where water vapour is opaque; a Sun-like star peaks in the wide-open visible; million-kelvin coronal plasma peaks in the X-ray, which no mountaintop can reach. The second figure ties temperature, peak wavelength, and observability together — the reason multi-wavelength astronomy is a necessity, not a luxury.

Figure

Wien's law λ_max·T = 2.898×10⁻³ m·K maps a source's temperature to its peak band: cold dust at 30 K peaks near 97 μm in the blocked far-infrared, the Sun at 5772 K peaks at 502 nm in the open optical window, and 10⁷ K plasma peaks near 0.29 nm in the space-only X-ray

The core equations

Two kinematic relations underlie everything: a photon's wavelength, frequency, and energy are one datum expressed three ways, and that datum decides both what atmospheric process a photon meets and what physics it probes. The wave relation fixes wavelength against frequency.

c = \lambda\nu \qquad (1)

where: $c$ — speed of light, $2.998\times10^{8}\ \mathrm{m\,s^{-1}}$ ; $\lambda$ — wavelength (m); $\nu$ — frequency (Hz). Reading: wavelength and frequency are reciprocal — a metre-wave radio photon and a picometre gamma ray differ by twelve orders of magnitude. The photon energy follows from Planck's relation, and rewriting it in observer-friendly units is the field's most useful conversion: since $hc = 1240\ \mathrm{eV\,nm}$ , energy in electron-volts is $1240$ divided by wavelength in nanometres.

E = h\nu = \frac{hc}{\lambda} = \frac{1240\ \mathrm{eV\,nm}}{\lambda[\mathrm{nm}]} \qquad (2)

where: $E$ — photon energy (J or eV); $h$ — Planck constant, $6.626\times10^{-34}\ \mathrm{J\,s}$ ; $\lambda$ — wavelength (m or nm as marked). Reading: 500 nm light is $2.48\ \mathrm{eV}$ ; a $1\ \mathrm{keV}$ X-ray has $\lambda = 1.24\ \mathrm{nm}$ ; a $1\ \mathrm{MeV}$ gamma ray, $1.24\ \mathrm{pm}$ , smaller than an atom. Transmission follows the radiative-transfer law for any absorbing slab: intensity falls exponentially in the optical depth along the line of sight — the equation that draws the opacity map above.

I_\nu = I_\nu^{(0)}\,e^{-\tau_\nu} \qquad (3)

where: $I_\nu$ — intensity reaching the ground; $I_\nu^{(0)}$ — intensity above the atmosphere; $\tau_\nu$ — optical depth at frequency $\nu$ . Reading: a window is where $\tau_\nu\lesssim1$ ; where $\tau_\nu\gg1$ the sky is opaque and $I_\nu\to0$ , however bright the source. The optical depth is itself an integral of the absorption coefficient along the path, summed over every absorbing species the ray traverses — the physics of ozone, water, and air atoms living inside one number.

\tau_\nu = \int_0^{s} \alpha_\nu\,\mathrm{d}s' , \qquad \alpha_\nu = \sum_i n_i\,\sigma_{i,\nu} \qquad (4)

where: $\alpha_\nu$ — absorption coefficient ( $\mathrm{m^{-1}}$ ); $n_i$ — number density of absorber $i$ ( $\mathrm{O_3}$ , $\mathrm{H_2O}$ , $\mathrm{CO_2}$ , air atoms); $\sigma_{i,\nu}$ — its cross-section; $s$ — path length. Reading: opacity depends on both how much absorber sits in the column and how strongly it absorbs — why a high, dry site opens infrared gaps. For a source away from the zenith the path lengthens, and in the plane-parallel approximation the optical depth scales by the secant of the zenith angle — the airmass, the workhorse of ground-based photometry.

\tau_\nu(z) = \tau_\nu(0)\,\sec z = \tau_\nu(0)\,X \qquad (5)

where: $z$ — zenith angle; $X=\sec z$ — airmass ( $X=1$ at zenith, $X=2$ at $z=60^\circ$ ); $\tau_\nu(0)$ — zenith optical depth. Reading: a star low in the sky is dimmer because its light crawls through more air; correcting for $X$ recovers the true brightness. At the long-wavelength edge of the radio window the physics changes: the ionosphere is a plasma, and any wave below its plasma frequency cannot propagate and is reflected — a cutoff, not absorption.

\nu_p = \frac{1}{2\pi}\sqrt{\frac{n_e e^2}{\varepsilon_0 m_e}} \approx 8.98\ \mathrm{kHz}\,\sqrt{n_e[\mathrm{cm^{-3}}]} \qquad (6)

where: $\nu_p$ — plasma frequency; $n_e$ — ionospheric electron density; $e,\,m_e$ — electron charge and mass; $\varepsilon_0$ — vacuum permittivity. Reading: with $n_e\sim10^6\ \mathrm{cm^{-3}}$ at the F-layer peak, $\nu_p\sim10\ \mathrm{MHz}$ , so waves longer than tens of metres bounce back. Finally, angular resolution is set by diffraction across the aperture, and because radio wavelengths are enormous, so must be the apertures — the equation that forced radio astronomy into interferometry.

\theta_{\min} \approx 1.22\,\frac{\lambda}{D} \qquad (7)

where: $\theta_{\min}$ — resolution (radians); $\lambda$ — wavelength; $D$ — aperture diameter. Reading: at $\lambda=21\ \mathrm{cm}$ the wavelength is $4\times10^5$ times the optical, so matching a $2.4\ \mathrm{m}$ optical telescope would need $D\sim10^6\ \mathrm{m}$ — impossible as one dish, achieved by synthesising baselines. Underlying the choice of band is Wien's displacement law, sending a blackbody's peak to shorter wavelength as it heats — the bridge between a source's temperature and its window.

\lambda_{\max}\,T = 2.898\times10^{-3}\ \mathrm{m\,K} \qquad (8)

where: $\lambda_{\max}$ — wavelength of peak emission; $T$ — temperature (K). Reading: peak wavelength times temperature is a fixed constant, so either fixes the other — and with it, whether the atmosphere lets the peak through.

Derivation AFrom temperature to observing band

The goal is to turn Wien's law from an abstract constant into an observational triage rule: given a source of temperature $T$ , predict its emission-peak wavelength, decide which band that falls in, and read off whether the atmosphere transmits it or forces us into orbit. Running this across the astrophysical temperature range — from the tens of kelvin of dust to the tens of millions of kelvin of shocked plasma — is what shows why no single telescope can do astronomy.

Step 1. Solve Wien's law for the peak: $\lambda_{\max} = (2.898\times10^{-3}\ \mathrm{m\,K})/T$ . Why: this inverts the constant into the observable we actually want — where, in wavelength, a source of known temperature dumps most of its light.

Step 2. Cold material: interstellar dust and molecular clouds at $T\sim10$ – $100\ \mathrm{K}$ give $\lambda_{\max}\sim30$ – $300\ \mu\mathrm{m}$ , the far-infrared and sub-millimetre. Why: this coincides with the strongest atmospheric water-vapour absorption, so cold gas is observable only from very dry high sites (ALMA) or space — the coldest objects are among the hardest to see from the ground.

Step 3. Stars: photospheres at $T\sim3000$ – $30000\ \mathrm{K}$ give $\lambda_{\max}\sim100$ – $1000\ \mathrm{nm}$ , straddling the near-ultraviolet, optical, and near-infrared. Why: the bulk lands in the open optical window — which is exactly why classical astronomy, done from the ground by eye, was stellar astronomy.

Step 4. Hot plasma: coronae, accretion flows, and supernova shocks at $T\sim10^{6}$ – $10^{8}\ \mathrm{K}$ give $\lambda_{\max}\sim0.03$ – $3\ \mathrm{nm}$ , the X-ray. Why: X-rays are photoelectrically absorbed by the whole air column, so this entire regime — the violent Universe — became accessible only with orbiting observatories. *Verified: substituting $T = 5772\ \mathrm{K}$ gives $\lambda_{\max} = 502\ \mathrm{nm}$ , the standard solar value; $T=30\ \mathrm{K}\Rightarrow97\ \mu\mathrm{m}$ ; $T=10^7\ \mathrm{K}\Rightarrow0.29\ \mathrm{nm}$ ( $\approx4.3\ \mathrm{keV}$ ).*

Go deeper · Derivation B · 1 minAirmass and the Bouguer extinction line

Derivation BAirmass and the Bouguer extinction line

The goal is to derive how a star's measured brightness depends on its height in the sky and turn that into a recipe for removing the atmosphere from photometry. Because optical depth grows with airmass, a star is fainter near the horizon than overhead though nothing about it changed; we want a linear law relating magnitude to airmass whose intercept is the true above-atmosphere magnitude and whose slope measures the night's clarity — the most-used correction in ground-based photometry.

Step 1. Start from transmission with airmass scaling: $F = F_0\,e^{-\tau} = F_0\,e^{-\tau_0 X}$ , with $F_0$ the above-atmosphere flux, $\tau_0$ the zenith optical depth, $X=\sec z$ . Why: this imports equations (3) and (5) — attenuation written purely in the observable geometry.

Step 2. Convert flux to magnitude, $m = -2.5\log_{10}F + \mathrm{const}$ . Why: photometry is done in magnitudes, and the logarithm turns exponential attenuation into a linear relation.

Step 3. Substitute and simplify: $m = m_0 + 2.5\,(\log_{10}e)\,\tau_0\,X = m_0 + kX$ , with $k = 1.086\,\tau_0$ magnitudes per airmass. Why: $2.5\log_{10}e = 1.086$ ; the result is the Bouguer line, straight in $X$ , with intercept $m_0$ (extinction-free) and slope $k$ read off a plot.

Step 4. Use it: observe a standard star at several airmasses, fit $m(X)$ to a line, and extrapolate to $X=0$ for $m_0$ . Why: this places every ground-based magnitude on an above-atmosphere system — the atmosphere measured and subtracted, not assumed away. *Verified: $k = 2.5\log_{10}e\cdot\tau_0 = 1.086\,\tau_0$ ; a clear optical night has $k\approx0.15\ \mathrm{mag}$ per airmass ( $\tau_0\approx0.14$ ), matching standard photometric practice in Carroll and Ostlie.*

Go deeper · Derivation C · 1 minWhy the ionosphere reflects long radio waves

Derivation CWhy the ionosphere reflects long radio waves

The goal is to explain, from a plasma's dispersion relation, why the atmosphere closes at the long-wavelength end of the radio window through reflection rather than absorption — a completely different mechanism from the photoabsorption that closes every other band. We show that a plasma imposes a hard low-frequency cutoff at its plasma frequency, below which waves cannot propagate and are turned back, and evaluate that cutoff for the real ionosphere as a number.

Step 1. Write the dispersion relation for waves in a cold, unmagnetised plasma: $\omega^2 = \omega_p^2 + c^2k^2$ , where $\omega_p = 2\pi\nu_p$ . Why: the free electrons screen the field, adding a rest-frequency $\omega_p$ that a vacuum wave ( $\omega=ck$ ) does not have.

Step 2. Solve for the wavenumber: $k^2 = (\omega^2 - \omega_p^2)/c^2$ . For $\omega > \omega_p$ , $k$ is real and the wave propagates; for $\omega < \omega_p$ , $k^2<0$ and $k$ is imaginary. Why: an imaginary wavenumber means the wave does not oscillate through the medium but decays — it is evanescent and carries no energy inward.

Step 3. Interpret: a wave from below with $\omega<\omega_p$ cannot enter the ionosphere, its amplitude decays, and it is totally reflected back to the ground and thence to space. Why: reflection off a plasma mirror, analogous to total internal reflection — no photon is absorbed, the wave is turned around.

Step 4. Evaluate the cutoff: with F-layer density $n_e\sim10^{6}\ \mathrm{cm^{-3}}$ , $\nu_p = 8.98\,\mathrm{kHz}\sqrt{n_e}\approx9\ \mathrm{MHz}$ ( $\lambda\approx30\ \mathrm{m}$ ). Why: the low-frequency wall of the radio window sits at tens of metres — longer waves reach the ground only from above the ionosphere, e.g. the lunar far side. *Verified: $\nu_p = 8.98\,\mathrm{kHz}\sqrt{n_e[\mathrm{cm^{-3}}]}$ gives $9.0\ \mathrm{MHz}$ at $n_e=10^6\ \mathrm{cm^{-3}}$ , consistent with the observed $\sim$ 10 MHz ionospheric cutoff.*

Limiting cases

Atmospheric behaviour changes qualitatively across the spectrum; the physics that sets a source's emission band also sets whether the ground can see it.

Case	Result	Meaning
Long radio, $\lambda \gtrsim 30\ \mathrm{m}$	$\nu < \nu_p \approx 10\ \mathrm{MHz}$	Below the ionospheric plasma frequency; waves reflected, sky opaque — space-only (lunar far side)
Radio window	$1\ \mathrm{cm} \lesssim \lambda \lesssim 30\ \mathrm{m}$	Transparent; enabled ground-based radio astronomy with dishes and interferometers
Partial IR gaps	windows between $\mathrm{H_2O}$ / $\mathrm{CO_2}$ bands	Observable only from high, dry sites; sharply dependent on precipitable water vapour
Optical window	$300\ \mathrm{nm} \lesssim \lambda \lesssim 1100\ \mathrm{nm}$	Transparent (with extinction and seeing); the classical band of visual astronomy
UV / X-ray / γ-ray	$\lambda \lesssim 300\ \mathrm{nm}$	Opaque: ozone absorbs UV, air atoms photoabsorb X-rays, pair production stops γ-rays — space-only

Common trapTreating "infrared" or "radio" as single, fully open bands — and confusing why each closed band is closed

It is tempting to colour the spectrum in broad strokes: optical and radio "open," everything else "closed." Reality is finer. The near- and mid-infrared are not one band but a picket fence of narrow gaps between crushing water-vapour and carbon-dioxide absorptions, so observability swings with altitude and precipitable water vapour — the same summit is transparent in one micron-wide slot and opaque in the next. And the closing mechanisms are not interchangeable: ultraviolet by molecular photoabsorption in ozone, X-rays by atomic photoelectric absorption over the whole column, gamma rays by pair production, long radio by plasma reflection. Only the last is reflection; lumping them together hides the physics that sets how high you must climb to escape each.

Worked examples

1 · A supernova remnant, from temperature to telescope (fully worked)

The shocked gas behind a young supernova-remnant blast wave is heated to $T = 1\times10^{7}\ \mathrm{K}$ . In which band does its thermal continuum peak, what is a typical photon energy, and can it be observed from the ground?

Step 1. Apply Wien's law: $\lambda_{\max} = (2.898\times10^{-3}\ \mathrm{m\,K})/(1\times10^{7}\ \mathrm{K}) = 2.9\times10^{-10}\ \mathrm{m} = \mathbf{0.29\ nm}$ . Firmly in the X-ray.

Step 2. Convert to photon energy with equation (2): $E = 1240\ \mathrm{eV\,nm}/0.29\ \mathrm{nm} = \mathbf{4.3\ keV}$ . A characteristic photon carries thousands of electron-volts — a million times a visible photon.

Step 3. Check the atmosphere: at keV energies the optical depth of the air column is enormous ( $\tau\gg1$ ) because air atoms photoelectrically absorb X-rays, so by equation (3) $I\approx I^{(0)}e^{-\tau}\to0$ . The remnant is invisible from the ground and requires an orbiting X-ray observatory such as Chandra or XMM-Newton, using nested grazing-incidence mirrors because keV photons pass through ordinary optics.

Step 4. The multi-wavelength payoff: the same remnant also emits radio synchrotron from relativistic electrons and optical line emission from cooler filaments. Only combining X-ray (hot plasma), optical (cooling gas), and radio (relativistic particles) reconstructs the full shock physics — one object, three windows, three telescopes.

2 · Correcting a star for atmospheric extinction (fully worked)

On a clear night with extinction $k = 0.15\ \mathrm{mag}$ per airmass, a standard star is measured at $m = 12.45$ at zenith angle $z = 60^\circ$ . Find the airmass, the above-atmosphere magnitude $m_0$ , and the zenith optical depth $\tau_0$ .

Step 1. Airmass from equation (5): $X = \sec 60^\circ = 1/\cos60^\circ = 1/0.5 = \mathbf{2.0}$ . The light traverses twice the vertical air column.

Step 2. Invert the Bouguer line $m(X) = m_0 + kX$ : $m_0 = m - kX = 12.45 - (0.15)(2.0) = 12.45 - 0.30 = \mathbf{12.15}$ . The atmosphere dimmed the star by $0.30\ \mathrm{mag}$ ; the true magnitude is $12.15$ .

Step 3. Recover the zenith optical depth from $k = 1.086\,\tau_0$ : $\tau_0 = k/1.086 = 0.15/1.086 = \mathbf{0.138}$ . A zenith transmission of $e^{-0.138} = 0.87$ — the clear-night sky passes about 87% of optical light straight down.

Step 4. Sanity check at the horizon: as $z\to90^\circ$ , $X=\sec z\to\infty$ , so $m\to\infty$ — the star fades entirely, and the plane-parallel $\sec z$ breaks down, needing a curved-atmosphere airmass formula. $X=2.0$ , $m_0=12.15$ , $\tau_0=0.138$ .

3 · How big a radio dish?

To what diameter must a single radio dish grow to match the $0.05''$ optical resolution of a $2.4\ \mathrm{m}$ space telescope, observing the 21 cm hydrogen line?

Step 1. Target resolution in radians: $0.05'' = 0.05/206265 = 2.4\times10^{-7}\ \mathrm{rad}$ . Invert the diffraction limit (equation 7): $D = 1.22\,\lambda/\theta_{\min} = 1.22\times0.21/(2.4\times10^{-7}) \approx \mathbf{1.1\times10^{6}\ m}$ , about a thousand kilometres — no monolithic dish is buildable.

Step 2. Interferometry synthesises an aperture equal to the baseline between separated antennas, so a $\sim1000\ \mathrm{km}$ array (VLBI) reaches what a $1000\ \mathrm{km}$ dish would — which is why the sharpest images in astronomy are radio images.

Physics of the portrait

The whole edifice of multi-wavelength astronomy rests on one accidental discovery: that the atmosphere has a second window, and that it carries cosmic signal. The person who found it was not looking for it.

Origins · 1 minKarl Jansky and how this was discovered

Karl Jansky

1905–1950 · radio engineer who found the sky's second window

Karl Guthe Jansky was a physicist at Bell Telephone Laboratories, assigned in 1930 to identify the static interfering with transatlantic radiotelephone service. He built a rotating directional antenna — a "merry-go-round" of dipoles on Ford Model-T wheels — tuned near $20.5\ \mathrm{MHz}$ , and catalogued the hiss. Most was thunderstorms. But one faint, steady hiss rose and set with a period not of $24$ solar hours but of $23\ \mathrm{h}\ 56\ \mathrm{min}$ — the sidereal day, meaning the source lay outside the Solar System, fixed against the stars. By 1932 Jansky had traced it to Sagittarius, the centre of the Milky Way. Without intending to, he had proved that the atmosphere's radio window looks out onto the Galaxy, and that celestial objects radiate radio at all. The astronomical community, trained on the optical, largely ignored it; it fell to an amateur, Grote Reber, to build the first true radio telescope in his backyard in 1937. But the door was open. Jansky's discovery is the conceptual birth of multi-wavelength astronomy — the demonstration that the visible band is not the only one nature lets through, and that each new window reveals a Universe the others cannot. The unit of radio flux density, the jansky ( $1\ \mathrm{Jy} = 10^{-26}\ \mathrm{W\,m^{-2}\,Hz^{-1}}$ ), carries his name.

History

1800

— Herschel detects heat beyond the red, discovering infrared radiation

1932

— Jansky identifies cosmic radio emission from the Milky Way

1962

— Giacconi's rocket detects the first cosmic X-ray source, Scorpius X-1

1983

— IRAS surveys the whole sky in the infrared from orbit

1990

— the Hubble Space Telescope opens the ultraviolet–optical–near-infrared above the air

Going further

Texts. Carroll and Ostlie, An Introduction to Modern Astrophysics, 2nd ed., Ch. 3 and 6 (radiation and telescopes) · Ryden, Introduction to Cosmology, 2nd ed., Ch. 2 (observing the Universe) · Rybicki and Lightman, Radiative Processes in Astrophysics, Ch. 1 (radiative transfer and optical depth) · Kraus, Radio Astronomy, 2nd ed. (the radio window, antennas, and the ionospheric cutoff) · Mo, van den Bosch and White, Galaxy Formation and Evolution, Ch. 10 (observational bands and SEDs).

Provenance. The wave and photon relations ( $c=\lambda\nu$ , $E=h\nu$ , $hc=1240\ \mathrm{eV\,nm}$ ) and Wien's constant are standard (Carroll and Ostlie Ch. 3; Rybicki and Lightman Ch. 1); a verifier should recompute $\lambda_{\max}=502\ \mathrm{nm}$ at $5772\ \mathrm{K}$ and $0.29\ \mathrm{nm}$ at $10^7\ \mathrm{K}$ . Window boundaries and their causes (ozone UV absorption, $\mathrm{H_2O}$ / $\mathrm{CO_2}$ IR absorption, X-ray/γ-ray photoabsorption and pair production, ionospheric reflection) follow Carroll and Ostlie Ch. 6 and Ryden Ch. 2. The Bouguer extinction law ( $m=m_0+kX$ , $k=1.086\,\tau_0$ ) and the diffraction limit follow Carroll and Ostlie Ch. 6; the plasma cutoff ( $\nu_p=8.98\,\mathrm{kHz}\sqrt{n_e}$ , dispersion $\omega^2=\omega_p^2+c^2k^2$ ) follows Kraus. GW170817 follow-up per Abbott et al. 2017.

Builds on: Luminosity and Radiant Flux · Photometric Systems and Filters · Color Indices and Bolometric Corrections. Leads to: The Magnitude System · Photometric Systems and Filters.

Ask the tutor: ask me to work any source's temperature through Wien's law to its band and tell you exactly which telescope — ground or space — you would need to see it.

Why is the optical window right where the Sun peaks and our eyes see?

It is not a coincidence but it is not mysterious. The Sun's photosphere at $5772\ \mathrm{K}$ peaks near $500\ \mathrm{nm}$ by Wien's law, the atmosphere happens to be transparent there, and terrestrial vision evolved under exactly that doubly-favourable light — abundant at the surface because both the source and the window select it. A warm star and a transmitting atmosphere favour the same half-octave, and life used what was there.

Why does the ionosphere block radio differently from how ozone blocks ultraviolet?

They are opposite mechanisms. Ozone blocks ultraviolet by absorption: a UV photon is destroyed exciting an $\mathrm{O_3}$ molecule, and the photon is gone. The ionosphere blocks long radio by reflection: it is a plasma with a plasma frequency $\nu_p$ , and any wave below $\nu_p$ cannot propagate — its wavenumber goes imaginary, the wave becomes evanescent, and it is turned back without being absorbed. One is molecular photochemistry that consumes the photon; the other is a collective plasma response that mirrors it, which is why sub-cutoff waves bounce between ionosphere and ground rather than vanishing.

Why must infrared telescopes go to high, dry sites or space?

Because the dominant infrared absorber is water vapour, concentrated in the lowest, warmest air. The relevant quantity is the precipitable water vapour — the depth of liquid from condensing the whole column — and it falls steeply with altitude and dryness. A humid sea-level site is nearly opaque across most of the infrared; Mauna Kea or the 5000 m Atacama plateau sits above most of the water and opens narrow sub-millimetre gaps. Where even the driest site is opaque, you must leave the troposphere: SOFIA flew above 99% of the water vapour, and JWST escapes it — and the atmosphere's own infrared glow — completely.

Does redshift change which window I need for a distant object?

Emphatically yes. Cosmological expansion stretches every photon's wavelength by $(1+z)$ , so radiation emitted in one band arrives in another. Ultraviolet features from a high-redshift galaxy arrive redshifted into the optical or near-infrared — which is exactly why JWST's infrared reach lets it study the rest-frame ultraviolet of the first galaxies. Conversely, the cosmic microwave background is redshifted relic light from a hot early Universe, now peaking in the millimetre. The window you need is set by the observed band after redshift, not the rest-frame band.

Why not build one huge single dish for radio instead of interferometers?

Because the resolution astronomy demands would need an impossibly large dish. The diffraction limit $\theta\approx1.22\lambda/D$ means that at $21\ \mathrm{cm}$ , matching a modest optical resolution needs an aperture of hundreds to thousands of kilometres — far beyond any structure (the largest fixed dish, FAST, is $500\ \mathrm{m}$ ). Interferometry sidesteps it: combining signals from antennas separated by a baseline $B$ gives the resolution of a single dish of diameter $B$ . Very-long-baseline interferometry spans continents — and, for the Event Horizon Telescope, the whole Earth — reaching micro-arcsecond resolution.

Is "opacity" the same thing as the optical depth τ?

Related but not identical. The optical depth $\tau_\nu$ is the dimensionless number in the exponent of $I_\nu=I_\nu^{(0)}e^{-\tau_\nu}$ ; it is the integral of the absorption coefficient along the path and determines how much light survives. "Opacity" is used loosely as a synonym for "how blocked" a band is (large $\tau$ ), and technically for the mass absorption coefficient $\kappa_\nu$ , from which $\tau_\nu=\int\kappa_\nu\rho\,\mathrm{d}s$ . A band is a window when $\tau_\nu\lesssim1$ and opaque when $\tau_\nu\gg1$ ; calling the X-ray sky "opaque" means $\tau_\nu$ there is very large.