Photometric Systems and Filters

A magnitude looks like a property of a star, but it is really a property of a measurement: a star seen through one specific window on the spectrum, weighted one specific way, referenced to one specific zero point. A photometric system is the machinery that makes that window reproducible. It fixes a set of filter passbands, each described by a response function $S(\lambda)$ ; it folds in the detector's sensitivity and the atmosphere's transparency; and it pins the zero of the scale with a network of standard stars. Change any of those and the number changes — which is why " $V = 12.4$ " is meaningful and "brightness $= 12.4$ " is not. Two conventions dominate. The Johnson–Cousins UBVRI system, born from photomultiplier astronomy in the 1950s, is a Vega system: the hot A0V star Vega is defined to sit at magnitude $\approx 0$ in every band, so a star's colours are measured relative to Vega's spectrum. The Sloan Digital Sky Survey's $ugriz$ system is built instead on the AB scale, where the zero point is tied to a fixed flux density in frequency, so an AB magnitude converts directly to $\mathrm{erg\,s^{-1}\,cm^{-2}\,Hz^{-1}}$ with no reference star at all. Everything downstream — colour indices as thermometers, bolometric corrections back to total luminosity, photometric redshifts, the calibration of Cepheids and Type Ia supernovae that anchors the cosmic distance ladder — rests on treating the passband as a known, fixed object. This article develops the response integral that turns a spectrum into a magnitude, the calibration chain that turns raw counts into a standard-system value, and the reason no two systems are exactly interchangeable.

Key takeaways

A photometric system is a standardized set of bandpasses, each defined by a response function $S(\lambda)$ (filter × detector × atmosphere), plus a network of standard stars that fix the zero points.
A band magnitude is the source spectrum integrated against that response, $m = -2.5\log_{10}\!\left(\int f_\lambda\,S(\lambda)\,\mathrm{d}\lambda\right) + \text{const}$ , so brightness depends on which wavelengths you count and how you weight them.
Johnson–Cousins UBVRI is Vega-referenced (Vega $\approx 0$ in every band); SDSS $ugriz$ is AB-referenced (zero point $= 3631\,\mathrm{Jy}$ ), so AB magnitudes are physical flux densities in disguise.
Colour indices such as $B-V$ or $g-r$ are magnitude differences; they measure the spectral slope and so track temperature, extinction, and metallicity.
Broadband, intermediate-band, and narrowband systems trade photon throughput for spectral resolution; the JHKLM infrared bands slot into the gaps between atmospheric water and $\mathrm{CO_2}$ absorption.
Converting between systems needs colour terms and airmass corrections — there is no exact universal conversion, since the bandpasses are physically different windows.
Key idea: a magnitude is not a property of a star alone but of a star viewed through a specific standardized bandpass with a defined zero point, so every photometric measurement is an integral of the source spectrum against a system response function.

Where this lives today

Photometric systems are the currency of every large survey. The Sloan Digital Sky Survey photometered hundreds of millions of objects in $ugriz$ on the AB scale, and its successors — the Dark Energy Survey, Pan-STARRS, and Rubin's LSST — push the same idea to billions of sources, where the limiting systematic is often not photon noise but the knowledge of each filter's $S(\lambda)$ and zero point to the millimagnitude. Space missions define their own systems and labour to tie them to the ground: Gaia's $G$ , $G_{\mathrm{BP}}$ , $G_{\mathrm{RP}}$ bands photometer nearly two billion stars, and JWST reports NIRCam and MIRI photometry on the AB scale out to the mid-infrared. Photometric redshifts — SED fitting across $ugrizJHK$ and beyond — are the only way to estimate distances for the faint galaxies that dominate deep imaging. And at the sharp end of cosmology, the calibration of Type Ia supernovae for the Hubble constant lives or dies on cross-instrument photometric consistency: the persistent $\sim 5\,\sigma$ tension between the local $H_0$ and the CMB value is, in part, an argument about zero points and colour terms carried across the distance ladder.

The physical picture

Point a filter at a star and you do not measure its spectrum; you measure one number — the flux that survives the transmission curve and lands on the detector. The transmission $S(\lambda)$ is a weighted window: near its peak almost every photon gets through, in its wings almost none. A blue filter and a red filter aimed at the same red giant report wildly different fluxes because they weight different parts of the same SED. The first figure shows a stellar spectrum sampled by the five Johnson–Cousins passbands, each labelled with its effective wavelength.

Figure

The UBVRI passbands as weighted windows on a stellar spectral energy distribution. Each band integrates the source flux fλ against its response S(λ); the labels give the standard effective wavelengths U 366 nm, B 438 nm, V 545 nm, R 641 nm, I 798 nm, and the measured magnitude in each band is m = −2.5 log₁₀(∫ fλ S dλ) + ZP

Because the two conventions weight the same spectrum against different standards, a single star carries two magnitudes, and the gap between them is a function of wavelength. The AB scale references a flat $f_\nu$ ; Vega is a hot star whose $f_\nu$ falls steeply toward the infrared, so it grows fainter against a flat standard with wavelength. The second figure is the number a survey astronomer keeps taped to the wall: the Vega-to-AB offset band by band, small and even sign-changing in the optical, climbing past $+1.8$ mag by the $K$ band.

Figure

The Vega-to-AB magnitude offset mAB − mVega as a function of band. It is near zero in the optical (U +0.79, B −0.09, V +0.02, R +0.21, I +0.45) but climbs steeply into the infrared (J +0.91, H +1.39, K +1.85) because Vega's flux density fν falls off toward long wavelengths while the AB system references a flat fν = 3631 Jy

The choice of how many bands, and how wide, is a throughput-versus-information trade. Broadband filters ( $\Delta\lambda/\lambda \sim 0.1$ ) gather photons greedily and reach the faintest objects, blurring every spectral feature together; narrowband filters (Strömgren $uvby$ , $\mathrm{H}\alpha$ ) isolate a single line or discontinuity but collect far fewer photons. The same logic tiles the whole spectrum: the JHKLM near-infrared bands sit not where an astronomer would ideally like them but where the atmosphere lets light through, in the windows between the water and $\mathrm{CO_2}$ absorption bands.

The core equations

The founding relation converts a spectrum into a single number. The detector adds up the source flux $f_\lambda$ weighted by the total system response $S(\lambda)$ , and the magnitude is $-2.5\log_{10}$ of that integral divided by the same integral over a reference spectrum, so the arbitrary units and the telescope aperture cancel.

m = -2.5\log_{10}\!\left(\frac{\int_0^\infty f_\lambda(\lambda)\,S(\lambda)\,\mathrm{d}\lambda}{\int_0^\infty f_\lambda^{\mathrm{ref}}(\lambda)\,S(\lambda)\,\mathrm{d}\lambda}\right) \qquad (1)

where: $m$ — apparent magnitude in the band; $f_\lambda$ — source flux density per unit wavelength ( $\mathrm{erg\,s^{-1}\,cm^{-2}\,\text{\AA}^{-1}}$ ); $S(\lambda)$ — total system response (filter × detector × atmosphere); $f_\lambda^{\mathrm{ref}}$ — reference-star flux density (Vega, for a Vega system). The ratio is a pure number, so any star with the reference's spectral shape lands at $m = 0$ . The AB system replaces the reference star with a reference flux density. Written in frequency, the response is weighted by $\mathrm{d}\nu/\nu$ (the natural measure for photon counting in log-frequency), and the additive constant $-48.60$ fixes the zero so that a flat source of $f_\nu = 3.631\times10^{-20}\,\mathrm{erg\,s^{-1}\,cm^{-2}\,Hz^{-1}}$ reads $m_{\mathrm{AB}} = 0$ in every band.

m_{\mathrm{AB}} = -2.5\log_{10}\!\left(\frac{\int f_\nu\,S(\nu)\,\mathrm{d}\nu/\nu}{\int S(\nu)\,\mathrm{d}\nu/\nu}\right) - 48.60 \qquad (2)

where: $f_\nu$ — flux density per unit frequency ( $\mathrm{erg\,s^{-1}\,cm^{-2}\,Hz^{-1}}$ ); $S(\nu)$ — response versus frequency; $-48.60 = -2.5\log_{10}(3.631\times10^{-20})$ . Because the zero point is a physical flux density, an AB magnitude inverts straight to $f_\nu = 3631\,\mathrm{Jy}\times 10^{-0.4\,m_{\mathrm{AB}}}$ with no reference star. A colour index is simply the difference of two band magnitudes for the same star. Because it is a difference, the distance and the overall brightness cancel, leaving a measure of spectral slope.

B-V,\quad U-B,\quad V-I,\quad g-r,\ \ldots \qquad (3)

where: $B-V$ — the workhorse index; larger (redder) means cooler; $U-B$ — sensitive to the Balmer jump; $g-r$ — the Sloan analogue. A colour is a temperature (and reddening) diagnostic precisely because it discards the flux normalization and keeps only the ratio of fluxes in two bands. The band does its weighting around a characteristic wavelength — the flux-weighted mean, called the effective wavelength. It depends weakly on the source spectrum, which is why a " $V$ -band" wavelength is quoted as a single number even though it shifts slightly for red versus blue stars.

\lambda_{\mathrm{eff}} = \frac{\int \lambda\, f_\lambda\, S(\lambda)\,\mathrm{d}\lambda}{\int f_\lambda\, S(\lambda)\,\mathrm{d}\lambda} \qquad (4)

where: $\lambda_{\mathrm{eff}}$ — flux-weighted mean wavelength of the band. It is where the band "sits" on the spectrum, the wavelength you attach to a colour when placing a point on an SED. Before magnitudes can be compared between telescopes, the atmosphere must be removed. Light entering at zenith angle $z$ traverses an airmass $X \approx \sec z$ , and each band dims by an extinction coefficient $k$ (magnitudes per airmass) measured nightly.

m_0 = m_{\mathrm{obs}} - k\,X, \qquad X \approx \sec z \qquad (5)

where: $m_0$ — magnitude above the atmosphere; $m_{\mathrm{obs}}$ — observed; $k$ — band extinction coefficient (larger in the blue, where Rayleigh scattering bites hardest); $X$ — airmass, $1$ at zenith. Plotting $m_{\mathrm{obs}}$ against $X$ gives a line of slope $k$ and intercept $m_0$ . Finally, an instrument's real bandpass never exactly matches the standard one, so the transformation to the standard system carries a linear colour term on top of a zero point and the extinction.

V = v - \zeta - \alpha\,(B-V) - k_V X \qquad (6)

where: $V$ — standard magnitude; $v$ — instrumental magnitude ( $-2.5\log_{10}$ of the count rate); $\zeta$ — instrumental zero point; $\alpha$ — colour term absorbing the bandpass mismatch; $(B-V)$ — standard colour. The coefficients $\zeta$ , $\alpha$ , $k_V$ are solved from standard stars spanning colour and airmass — that fit is joining a photometric system.

Derivation AFrom photons in the aperture to a band magnitude

The goal is to build equation (1) from first principles for a real detector, and to expose a subtlety the compact formula hides: a CCD counts photons, not energy, so the response in the integral is not the energy-weighted transmission but a photon-weighted one, carrying an extra factor of wavelength. We start from a single photon's energy, convert an energy flux into a photon arrival rate, integrate over the passband, and only then take the logarithm.

Step 1. A photon of wavelength $\lambda$ carries energy $hc/\lambda$ , so an energy flux density $f_\lambda$ corresponds to a photon flux density $f_\lambda\,\lambda/(hc)$ — more photons per erg toward the red. Why: the detector responds to photon arrivals, so the extra factor of $\lambda$ enters.

Step 2. Integrate the photon rate over the band, weighting by the response $S(\lambda)$ (the product of filter, atmosphere, and quantum efficiency): the count rate is $N \propto \int f_\lambda\,S(\lambda)\,\dfrac{\lambda}{hc}\,\mathrm{d}\lambda$ . Why: the total signal sums the photons that got through, and $h$ , $c$ fold into the zero point.

Step 3. Take the instrumental magnitude $m = -2.5\log_{10} N + C$ , where $C$ collects aperture, exposure time, and the $hc$ factor. Why: a multiplicative factor inside $N$ becomes an additive shift after the logarithm — exactly what a zero point removes.

Step 4. Fix $C$ with standard stars of known magnitude; the result is equation (1), with the reference integral playing the role of $C$ . Why: no telescope can predict its own zero point from the spectrum alone — calibrators supply it.

Go deeper · Derivation B · 1 minThe instrumental-to-standard calibration chain

Derivation BThe instrumental-to-standard calibration chain

The goal is to derive the working equation of every night's photometric reduction — the linear relation between what the CCD records and the magnitude a colleague on another telescope would report — by peeling off, one at a time, the three things that separate them: the atmosphere overhead, the arbitrary sensitivity of this instrument, and the fact that its bandpass is not quite the standard one. We begin with the raw count rate and end with a formula linear in airmass and colour.

Step 1. Define the instrumental magnitude from the raw signal, $m_{\mathrm{inst}} = -2.5\log_{10}(N/t)$ , where $N$ is the total electron count and $t$ the exposure time. Why: dividing by $t$ makes exposures of different length comparable.

Step 2. Correct for the atmosphere: the observed value dims by $kX$ , so $m_{\mathrm{inst,0}} = m_{\mathrm{inst}} - kX$ . Why: the path length scales as airmass $X\approx\sec z$ , and each unit removes $k$ magnitudes that must be undone before sites can agree.

Step 3. Relate the above-atmosphere value to the standard magnitude through a zero point and colour term, $m_{\mathrm{inst,0}} = m_{\mathrm{std}} - \zeta - \alpha C$ . Why: the instrument's response differs from the standard, and to first order that difference depends on the star's colour $C$ .

Step 4. Combine and solve: $m_{\mathrm{std}} = m_{\mathrm{inst}} + \zeta + \alpha C - kX$ , with $\zeta$ , $\alpha$ , $k$ fit from standard stars spanning colour and airmass. Why: three unknowns need a spread of calibrators, and that joint fit is what "putting data on a standard system" means.

Go deeper · Derivation C · 1 minWhy the Vega and AB magnitudes of a star differ by a fixed per-band offset

Derivation CWhy the Vega and AB magnitudes of a star differ by a fixed per-band offset

The goal is to show that although the Vega and AB systems integrate the same source flux through the same passband, their magnitudes differ by a constant that depends only on the band, not the star — the ratio of the two reference fluxes. That is what lets a single tabulated offset convert an entire catalogue, and it explains why the offset is tiny in the optical but large in the infrared. We write each magnitude against its own reference, subtract, and watch the source flux cancel.

Step 1. Write the Vega magnitude as a ratio against Vega's in-band flux, $m_{\mathrm{Vega}} = -2.5\log_{10}(F_{\mathrm{band}}/F_{\mathrm{band}}^{\mathrm{Vega}})$ , so Vega gives $m\approx 0$ . Why: the Vega system declares one star's magnitude zero, fixing the denominator.

Step 2. Write the AB magnitude against the flat- $f_\nu$ reference, $m_{\mathrm{AB}} = -2.5\log_{10}(F_{\mathrm{band}}/F_{\mathrm{band}}^{\mathrm{AB0}})$ , with $F_{\mathrm{band}}^{\mathrm{AB0}}$ the in-band flux at $f_\nu = 3.631\times10^{-20}\,\mathrm{erg\,s^{-1}\,cm^{-2}\,Hz^{-1}}$ . Why: fixing the denominator with a flux density is what makes AB invertible to physical units.

Step 3. Subtract: $m_{\mathrm{AB}} - m_{\mathrm{Vega}} = -2.5\log_{10}(f_\nu^{\mathrm{Vega}}/3.631\times10^{-20})$ ; the source flux $F_{\mathrm{band}}$ cancels. Why: both systems weight the identical source by the identical response, so everything about the star divides out — the offset is a pure property of the band.

Step 4. Read the numbers: in the optical Vega's $f_\nu$ is close to $3631\,\mathrm{Jy}$ , so the offset is small and can flip sign ( $U\approx+0.79$ , $B\approx-0.09$ , $V\approx+0.02$ ); toward the infrared it climbs ( $J\approx+0.91$ , $K\approx+1.85$ ). Why: Vega's $f_\nu$ falls steeply with wavelength, so it grows dim against a flat standard in the infrared. *The $K$ -band value follows directly from Vega's in-band flux density $f_\nu^{\mathrm{Vega}}(K)\approx 665\,\mathrm{Jy}$ : $-2.5\log_{10}(665/3631) = +1.84$ .*

Limiting cases

The response integral collapses to something simpler in each of the idealized regimes below.

Case	Result	Meaning
Monochromatic / narrowband	$m \approx -2.5\log_{10} f_{\lambda_0} + C$	when $S(\lambda)$ is sharply peaked at $\lambda_0$ , the magnitude tracks the flux density at that one wavelength — narrowband filters approach this
Flat spectrum in AB	$m_{\mathrm{AB}} = -2.5\log_{10} f_\nu - 48.60$	for constant $f_\nu$ the AB magnitude is band-independent by construction, so equal AB magnitudes in all bands means a flat SED
Vega reference	$m_{\mathrm{Vega}} = -2.5\log_{10}\!\left(\int f_\lambda S\,\mathrm{d}\lambda\big/\int f_\lambda^{\mathrm{Vega}} S\,\mathrm{d}\lambda\right)$	Vega magnitudes are ratios against Vega's spectrum, so Vega has $m\approx 0$ and colours $\approx 0$ in every band
Rayleigh–Jeans tail	$f_\lambda \propto \lambda^{-4}$	on a hot star's long-wavelength tail the SED is a fixed power law, so colours there depend weakly on temperature
Wide, featureless band	$m \to -2.5\log_{10}\!\big(\bar f_\lambda \!\int S\,\mathrm{d}\lambda\big)+C$	for a smooth source, only the mean flux and integrated throughput matter

Common trapTreating magnitudes from different systems as interchangeable

The single most common photometric error is writing " $V = g$ " or quoting a Vega magnitude where an AB magnitude is expected. They are not the same number. Different systems use physically different bandpasses — Johnson $V$ and Sloan $g$ peak at different wavelengths and widths — so they integrate different slices of the spectrum, and their zero-point conventions differ too. A naive equation between them can be wrong by several tenths of a magnitude, and the error is colour-dependent, so it does not cancel in a difference. Converting requires a colour term and, for Vega-versus-AB, the per-band offset above ( $V$ shifts by $+0.02$ , but $K$ by $+1.85$ ). Always state the system and reference; a magnitude without them is not a measurement.

Worked examples

1 · From an AB magnitude to a physical flux and a photon rate (fully worked)

A galaxy is measured at $g_{\mathrm{AB}} = 20.0$ on a $2.5\,\mathrm{m}$ telescope. The $g$ band is centred near $475\,\mathrm{nm}$ with an effective width $\Delta\lambda \approx 140\,\mathrm{nm}$ . Find its flux density $f_\nu$ , and estimate the photon rate the telescope collects.

Step 1. Invert the AB definition. Since the zero point corresponds to $3631\,\mathrm{Jy}$ , $f_\nu = 3631\,\mathrm{Jy}\times 10^{-0.4\,m_{\mathrm{AB}}} = 3631\times 10^{-8}\,\mathrm{Jy} = \mathbf{36.3\ \mu Jy} = 3.63\times10^{-28}\,\mathrm{erg\,s^{-1}\,cm^{-2}\,Hz^{-1}}$ . This is the point of the AB system: magnitude straight to physical flux density, no reference star.

Step 2. Convert to $f_\lambda = f_\nu\,c/\lambda^2$ to fold against the band width. With $\lambda = 4750\,\text{\AA}$ , $f_\lambda \approx 4.8\times10^{-17}\,\mathrm{erg\,s^{-1}\,cm^{-2}\,\text{\AA}^{-1}}$ ; over $\Delta\lambda = 1400\,\text{\AA}$ the energy flux is $F \approx 6.8\times10^{-14}\,\mathrm{erg\,s^{-1}\,cm^{-2}}$ .

Step 3. Multiply by a $2.5\,\mathrm{m}$ collecting area, $A = \pi(125\,\mathrm{cm})^2 \approx 4.9\times10^{4}\,\mathrm{cm^2}$ , giving $\approx 3.3\times10^{-9}\,\mathrm{erg\,s^{-1}}$ . Each $4750\,\text{\AA}$ photon carries $hc/\lambda \approx 4.2\times10^{-12}\,\mathrm{erg}$ , so the rate is $\approx \mathbf{8\times10^{2}\ photons\ s^{-1}}$ before losses. A $g = 20$ galaxy delivers $f_\nu \approx 36\ \mu$ Jy and roughly a thousand photons per second to a 2.5 m mirror — why deep surveys reach far fainter.

2 · Reddening from a colour excess

A G0V star (intrinsic colour $(B-V)_0 = 0.59$ ) is observed at $B-V = 0.79$ . Find the colour excess, the $V$ -band extinction, and the error in distance if extinction is ignored.

Step 1. The colour excess is the reddening above the intrinsic colour: $E(B-V) = (B-V) - (B-V)_0 = 0.79 - 0.59 = \mathbf{0.20}$ . Dust absorbs blue light more than red, so it always reddens a star relative to its spectral type.

Step 2. Apply the standard Galactic extinction ratio $R_V \approx 3.1$ : $A_V = R_V\,E(B-V) = 3.1\times 0.20 = \mathbf{0.62\ mag}$ of dimming in $V$ .

Step 3. Propagate to distance. Since $m - M = 5\log_{10}(d/10\,\mathrm{pc}) + A_V$ , ignoring $A_V$ attributes the extra $0.62$ mag to distance, overestimating it by $10^{A_V/5} = 10^{0.124} = \mathbf{1.33}$ . A modest $E(B-V) = 0.20$ inflates the inferred distance by 33 %, which is why extinction correction is not optional on the distance ladder.

3 · Vega-to-AB conversion in the K band

A star has $K = 10.00$ on the Vega system. What is its AB magnitude?

Step 1. Use the per-band offset from the third derivation with Vega's in-band flux $f_\nu^{\mathrm{Vega}}(K)\approx 665\,\mathrm{Jy}$ : $m_{\mathrm{AB}} - m_{\mathrm{Vega}} = -2.5\log_{10}(665/3631) = \mathbf{+1.84}$ .

Step 2. Add: $K_{\mathrm{AB}} = 10.00 + 1.84 = \mathbf{11.84}$ . The same star is nearly two magnitudes fainter on the AB scale in $K$ — far too large to ignore when combining infrared and optical catalogues.

Physics of the portrait

The photometric system is where measurement discipline meets stellar physics: a set of conventions that let one observer's number mean the same thing on another's telescope, a century after the eye gave way to the photomultiplier and then the CCD.

Origins · 1 minHarold Johnson and how this was discovered

Harold Johnson

1921–1980 · architect of the UBV photoelectric system

Harold Lester Johnson was the observer who turned photometry from an art into a system. In the early 1950s the field was a patchwork of photographic and visual estimates that no two observatories could reconcile to better than a few tenths of a magnitude. Working with William Wilson Morgan, Johnson exploited the newly practical photomultiplier tube — which counted photoelectrons linearly, unlike the treacherous photographic plate — to define, in 1953, the $UBV$ system: three specified passbands, a network of standard stars, and a zero point tied to A0V stars like Vega. The choice of bands was physical: $U$ straddling the Balmer jump, $B$ and $V$ bracketing it, so that $U-B$ and $B-V$ became a diagnostic of temperature and surface gravity. He extended it into the infrared as detectors allowed, and Alan Cousins later refined the red and near-infrared bands into the Johnson–Cousins system still in use. His insistence that a magnitude is meaningless without a fully specified response function and a standard-star network is the backbone of everything here — and of every survey that followed. The AB system of Oke and Gunn and the SDSS $ugriz$ system are later answers to the question Johnson posed: how do you make brightness a number everyone can trust?

History

1856

— Pogson fixes the magnitude scale at a hundredfold brightness ratio per five magnitudes

1953

— Johnson and Morgan define the UBV photoelectric system and its standard stars

1976

— Cousins refines the R and I bands into the Kron–Cousins system

1983

— Oke and Gunn introduce the AB magnitude scale tied to a physical flux density

1996

— the SDSS ugriz system is specified on the AB scale by Fukugita and collaborators

Going further

Texts. Carroll & Ostlie, An Introduction to Modern Astrophysics, 2nd ed., §3.6 (photometric systems and colour indices) · Bessell, M. S. 2005, ARA&A 43, 293 (Standard Photometric Systems — the definitive review) · Fukugita et al. 1996, AJ 111, 1748 (the SDSS $ugriz$ system) · Oke & Gunn 1983, ApJ 266, 713 (the AB scale) · Binney & Tremaine, Galactic Dynamics, 2nd ed., Appendix on photometry · Mo, van den Bosch & White, Galaxy Formation and Evolution, Ch. 2 (observational foundations). The AB zero point $f_\nu = 3.631\times10^{-20}\,\mathrm{erg\,s^{-1}\,cm^{-2}\,Hz^{-1}} = 3631\,\mathrm{Jy}$ and the constant $-48.60$ are due to Oke & Gunn (1983); the $ugriz$ specification is from Fukugita et al. (1996); the extinction ratio $R_V \approx 3.1$ is the standard diffuse-ISM value. Key numerical checks: $m_{\mathrm{AB}} = 20 \Rightarrow f_\nu = 36.3\,\mu\mathrm{Jy}$ and $\approx 8\times10^2$ photons s $^{-1}$ for a $2.5\,\mathrm{m}$ aperture; $E(B-V)=0.20 \Rightarrow A_V = 0.62$ and a $33\%$ distance error; the $K$ -band offset $-2.5\log_{10}(665/3631) = +1.84$ .

Prerequisites. Luminosity and Radiant Flux · The Magnitude System · Optical Telescopes and Their Optics. Leads to. Colour Indices and Bolometric Corrections · Absolute Magnitude and the Distance Modulus · Standard Candles · The Cosmic Distance Ladder.

Ask the tutor: Want to see how these same bandpasses turn into the colour–magnitude diagram of a star cluster and date it — or how AB zero points propagate to the Hubble tension?

Why does the same star have two different magnitudes on the Vega and AB systems?

Because the two systems answer "how bright is zero?" differently. The Vega system declares the star Vega to be magnitude zero in every band, so every star is measured as a ratio against Vega's spectrum. The AB system instead declares a fixed flux density — $3631\,\mathrm{Jy}$ , flat in frequency — to be zero. Since Vega's spectrum is not flat in $f_\nu$ , the two zero points disagree by an amount that depends on where the band sits: only $+0.02$ mag in $V$ , but $+1.85$ mag in $K$ . The star's actual flux is identical; only the yardstick's zero mark has moved — which is why you must always state which system a magnitude is on.

What exactly is the system response S(λ), and why does the atmosphere belong in it?

$S(\lambda)$ is the total probability that a photon of wavelength $\lambda$ arriving at the top of the atmosphere ends up recorded as signal — the product of atmospheric transmission, mirror reflectivity, filter transmission, and detector quantum efficiency. The atmosphere belongs because it is genuinely part of what you measured, and it changes with airmass and from night to night. Reductions correct to "above the atmosphere" using the nightly extinction coefficient $k$ , but strictly the ground-based bandpass is atmosphere-included — one reason space photometry defines cleaner, more stable systems.

Why weight by photon count instead of energy for a CCD?

Because a CCD, like the eye and the photomultiplier before it, registers photon arrivals, not the energy they carry. A red photon and a blue photon each liberate (roughly) one photoelectron, though the blue one carries more energy. So the count rate is the integral of the photon flux — the energy flux $f_\lambda$ times $\lambda/(hc)$ , an extra factor of wavelength. For broad bands this shifts the effective wavelength slightly to the red and matters at the percent level, and it is why the exact definition of a band depends on the detector class.

How does a colour index measure temperature if it throws away the total brightness?

That is exactly why it works. Total brightness depends on distance, size, and dust — none of which you know a priori. A colour index is a ratio of fluxes in two bands, so distance and overall scale cancel, leaving only the shape of the spectrum. For a rough blackbody, that shape is set by temperature: a hot star radiates relatively more in the blue, so $B-V$ is small (even negative); a cool star radiates more in the red, so $B-V$ is large. The relation is monotonic over most of the main sequence, which makes the colour–magnitude diagram a temperature axis. The catch is reddening, which also increases $B-V$ — so a colour is a thermometer only after the dust is accounted for.

Why not just take a spectrum instead of measuring a few bands?

Photons. A spectrograph spreads a star's light across thousands of resolution elements, so each receives a tiny fraction of the flux; photometry dumps all the band's light into one measurement. For a faint galaxy at the survey limit, a usable spectrum might need hours while broadband photometry needs seconds, and you can photometer millions of objects at once across a wide field. That is why photometric redshifts — SED shapes from a handful of bands — exist at all: they are the only way to get distances for the billions of sources too faint to spectroscope. Spectroscopy is the gold standard for precision; photometry is what makes whole-sky surveys feasible.

What is the airmass and why does it dim a star?

The airmass $X$ is the amount of atmosphere the light traverses relative to straight up, $X \approx \sec z$ for zenith angle $z$ ( $X = 1$ at the zenith, $\approx 2$ at $60^\circ$ from it). The more atmosphere in the path, the more the light is scattered and absorbed, so a star looks fainter the lower it sits — and more so in the blue, which also reddens it. Reductions track a star as it rises and sets, fit the observed magnitude as a line in $X$ , and read off the extinction coefficient $k$ (slope) and the above-atmosphere magnitude (intercept). Ignoring airmass introduces a colour- and time-dependent error.