PX282 - A6 - magnitudes, band passes, colours

see: PX158 - D2 - brightness and magnitudes

magnitude

astronomy uses magnitude to measure the brightness
it is on a logarithmic scale
originally, it was designed to mimic human eye response

apparent magnitude

it is the magnitude as seen from a distance, $d$
for two stars with apparent magnitudes, $m_{1}$ and $m_{2}$ , and fluxes, $f_{1}$ and $f_{2} :$

\begin{matrix} (1) & m_{1} - m_{2} = - 2.5 \log (\frac{f_{1}}{f_{2}}) \end{matrix}

the scale needs a zero-point, which is taken to be vega, $m_{v e g a} = 0 \forall λ$
therefore, the apparent magnitude for any star is:

m_{*} = - 2.5 \log (\frac{f_{*}}{f_{v e g a}})

this scale makes numbers easy to work with, eg:
- sirius, $m = - 1.4$
- sun, $m_{⊙} = - 2.7$
- proxima cententari, $m = 11$
note: larger the apparent magnitude, the fainter a fainter star
$\sim 5 m a g n i t u d e s s \approx 100 \times b r i g h t n e s s$
flux and magnitude are so far bolometric*, ie. measured across all wavelengths, or formally:

f = \int_{0}^{\infty} f_{λ} d λ

where, $f_{λ}$ is the flux density $[J s^{- 1} m^{- 2} n m^{- 1}]$

in practice, all wavelengths cannot be measured
$λ -$ ranges are selected using filters
a filter has a transmission function, $T_{A} :$

{\bar{f}}_{A} = \frac{\int_{0}^{\infty} T_{λ} f_{λ} d λ}{\int_{0}^{\infty} T_{λ} d λ}

eg:
- $90 %$ at $λ$

colours and temperature

image; speclite developers

$m$ in a bandpass: $m_{B}, m_{U}, m_{V}$
- $U - B colour = m_{U} - m_{B}$
- $B - V colour = m_{B} - m_{V}$
$B$ has shorter $λ$ than $V$ , so a star with a $B - V < 0$ , has $m_{B} < m_{V}$ , is 'bluer' than vega ( $m_{v e g a} = 0$ in all bands)
colour is a strong function of temperature

absolute magnitude

it says how bright would a star would be if it were at a distance of $10 p c$ , ie: the magnitude of a star at a distance of $10 p c$
therefore, $M = m (at 10 pc)$
flux has an inverse square relation with the distance, ie: $\frac{f (10 p c)}{f_{d}} = {(\frac{d}{10 p c})}^{2}$
using equation $(1) :$

\begin{aligned} M - m & = - 2.5 \log (\frac{f (10 p c)}{f_{d}}) \\ = - 2.5 \log {(\frac{d}{10 p c})}^{2} \\ ∴ M - m & = - 5 \log (\frac{d}{10 p c}) \end{aligned}

if $L ↑$ then $M ↓$ (brighter)