Photometry toolkit

Pogson's relation and its inverse:

Magnitudes of unresolved stars do not add; their fluxes do:

Going backwards, with a luminosity ratio $L_2/L_1$ as in the 2023 eclipsing binaries,

Dust dims and reddens. The colour excess is $E(B-V) = (B-V) - (B-V)_0$, with the intrinsic colour from the spectral type and luminosity class, and the extinction is

in the diffuse Galactic ISM. Since $E_{X-V} = A_X - A_V$ and $A_X\to0$ at long wavelengths, plotting $E_{X-V}/E_{B-V}$ against $1/\lambda$ and extrapolating to $1/\lambda = 0$ gives $-R_V$ (2015 P2).

If a source is seen in two bands with known absolute magnitudes, the difference of the two apparent distance moduli is a colour excess: $\mu_R - \mu_I = A_R - A_I = E(R-I)$.

The bolometric correction converts a band magnitude to the total: $m_{\rm bol} = m_V + BC$, with BC from a table against temperature (most negative for very hot and very cool stars). Then

and $L = 4\pi R^2\sigma T^4$ links luminosity, radius and temperature. Temperatures come from colours through a calibration curve, which is how the 2015 Baade–Wesselink problem gets $T$ from $V-R$.

The AB system is defined by a flat $f_\nu$:

Converting between frequency and wavelength densities uses $f_\nu\,|d\nu| = f_\lambda\,|d\lambda|$, so $f_\lambda = f_\nu c/\lambda^2$. A source with constant $m_{\rm AB}$ therefore has $f_\lambda\propto\lambda^{-2}$ ($\beta = -2$).

For distant galaxies, observed wavelengths are $(1+z)$ times the rest-frame ones. Track every $(1+z)$ factor explicitly: in 2018 D1.4 one missing factor of 7.6 changes the conclusion.

When the extinction of each star is unknown, combine two bands so that it cancels. The Wesenheit magnitude

is unchanged by reddening that follows the assumed law: dimming moves $V$ and $V-I$ together exactly so that $W$ stays fixed. In 2017 D1 the Wesenheit PL relation gives LMC distances with a scatter of 0.9 kpc against 4.1 kpc in V.

The surface brightness $S_V = m_V + 5\log\theta$ is a similar trick. It depends almost only on temperature, and therefore on colour, so a colour predicts the angular diameter.

From the ground, $m = m_0 + kX$, with airmass $X\approx\sec z$ for zenith distance $z<60^\circ$. Plot $m$ against $X$ for one star over a night: the slope is the extinction coefficient $k$ and the intercept at $X = 0$ is the magnitude above the atmosphere.