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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?

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  • $\begingroup$ @uhoh erg. sec^{-1}. Hz ^{-1} $\endgroup$
    – Arpan Das
    Commented Mar 5, 2020 at 18:18
  • $\begingroup$ Of course M's are unit less. They are absolute magnitude. You asked me the unit of luminosity $\endgroup$
    – Arpan Das
    Commented Mar 6, 2020 at 21:03
  • $\begingroup$ Oh, I didn't notice my previous moment. Does this look okay? $\endgroup$
    – uhoh
    Commented Mar 6, 2020 at 22:06
  • $\begingroup$ 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). $\endgroup$
    – pela
    Commented Dec 16, 2021 at 16:13

1 Answer 1

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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.

If you want Vega magnitudes or Johnson magnitudes or whatever, you use a slightly different zeropoint. But you don't want that, just stick with the AB magnitudes.

Or you just use astropy's tools for converting between fluxes and magnitudes.

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    $\begingroup$ I like this explanation. $\endgroup$ Commented Mar 10, 2020 at 18:54
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    $\begingroup$ and so do I :-) $\endgroup$
    – uhoh
    Commented Mar 11, 2020 at 0:59

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