# How to convert luminosity at rest frame wavelength of 1450 A to absolute magnitude at 1450 A?

How do I convert the luminosity (erg sec$$^{-1}$$ Hz$$^{-1}$$) of a quasar at a rest frame wavelength of 1450 Angstroms to absolute magnitude at the same wavelength?

I know that the bolometric luminosity is related to absolute magnitude via this relation from Bolometric magnitude:

$$M_{bol, ★} - M_{bol, ☉} = -2.5 log_{10} \left( \frac{L_★}{L_☉} \right)$$

Should the same relation be valid for the 1450 A rest frame luminosity and I just used that luminosity instead of the bolometric luminosity?

• @uhoh erg. sec^{-1}. Hz ^{-1} Commented Mar 5, 2020 at 18:18
• Of course M's are unit less. They are absolute magnitude. You asked me the unit of luminosity Commented Mar 6, 2020 at 21:03
• Oh, I didn't notice my previous moment. Does this look okay?
– uhoh
Commented Mar 6, 2020 at 22:06
• Hi Arjan, did my answer help you? If so, I'd be grateful if you could mark it as an accepted answer (also to help future readers).
– pela
Commented Dec 16, 2021 at 16:13

The absolute magnitude of an object is defined as the brightness of the object observed at a distance of $$d = 10\,\mathrm{pc}$$. With this distance, you can convert the luminosity density $$L_\nu$$ in $$\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{Hz}^{-1}$$ to a flux density $$f_\nu$$ in $$\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}\,\mathrm{Hz}^{-1}$$: $$f_\nu = \frac{L_\nu}{4\pi \times (10\,\mathrm{pc})^2}.$$
From there, you use the definition of the AB magnitude from Oke & Gunn (1983): $$M_\mathrm{AB} = -2.5\log f_\nu \,\,–\,\, 48.60.$$ Note that there's an error in the original equation, as the authors write plus 48.60 instead of minus.