<p>The absolute magnitude of an object is defined as the brightness of the object observed at a distance of <span class="math-container">$d = 10\,\mathrm{pc}$</span>. With this distance, you can convert the luminosity density <span class="math-container">$L_\nu$</span> in <span class="math-container">$\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{Hz}^{-1}$</span> to a flux density <span class="math-container">$f_\nu$</span> in <span class="math-container">$\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}\,\mathrm{Hz}^{-1}$</span>:
<span class="math-container">$$
f_\nu = \frac{L_\nu}{4\pi \times (10\,\mathrm{pc})^2}.
$$</span></p>

<p>From there, you use the definition of the AB magnitude from <a href="https://ui.adsabs.harvard.edu/abs/1983ApJ...266..713O" rel="nofollow noreferrer">Oke &amp; Gunn (1983)</a>:
<span class="math-container">$$
M_\mathrm{AB} = -2.5\log f_\nu \,\,–\,\, 48.60.
$$</span>
Note that there's an error in the original equation, as the authors write <em>plus</em> 48.60 instead of minus.</p>

<p>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.</p>

<p>Or you just use <a href="https://docs.astropy.org/en/stable/units/logarithmic_units.html" rel="nofollow noreferrer">astropy's tools for converting between fluxes and magnitudes</a>.</p>

CC BY-SA 4.0
2020-03-10T18:54:47.850Z
I like this explanation.
Kornpob Bhirombhakdi
CC BY-SA 4.0
2020-03-11T00:59:34.083Z
and so do I :-)
uhoh
CC BY-SA 4.0
2020-03-08T22:23:34.827Z
2020-03-08T22:23:34.827Z
pela
<p>How do I convert the luminosity (erg sec<span class="math-container">$^{-1}$</span> Hz<span class="math-container">$^{-1}$</span>) of a quasar at a rest frame wavelength of 1450 Angstroms to absolute magnitude at the same wavelength? </p>

<p>I know that the bolometric luminosity is related to absolute magnitude via this relation from <a href="https://en.wikipedia.org/wiki/Absolute_magnitude#Bolometric_magnitude" rel="nofollow noreferrer">Bolometric magnitude</a>:</p>

<p><span class="math-container">$$M_{bol, ★} - M_{bol, ☉} = -2.5 log_{10} \left( \frac{L_★}{L_☉} \right)$$</span></p>

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

CC BY-SA 4.0
2020-03-05T18:18:13.273Z
@uhoh erg. sec^{-1}. Hz ^{-1}
Arpan Das
CC BY-SA 4.0
2020-03-06T21:03:07.727Z
Of course M's are unit less. They are absolute magnitude. You asked me the unit of luminosity
Arpan Das
CC BY-SA 4.0
2020-03-06T22:06:20.467Z
Oh, I didn't notice my previous moment. Does this look okay?
uhoh
CC BY-SA 4.0
2021-12-16T16:13:54.070Z
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
<luminosity><photometry><quasars><absolute-magnitude>
How to convert luminosity at rest frame wavelength of 1450 A to absolute magnitude at 1450 A?
Arpan Das
# How to convert luminosity at rest frame wavelength of 1450 A to absolute magnitude at 1450 A?

(Asked by: username_0 on 2020-03-03)

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](https://en.wikipedia.org/wiki/Absolute_magnitude#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?

**Tags:** luminosity, photometry, quasars, absolute-magnitude
> **Comments:**
> **username_0**: @uhoh erg. sec^{-1}. Hz ^{-1}  
> **username_0**: Of course M's are unit less. They are absolute magnitude. You asked me the unit of luminosity  
> **username_1**: Oh, I didn't notice my previous moment. Does this look okay?  
> **username_2**: 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).  

---

## 1 answers

### Answer by username_2 on 2020-03-08. Score: 4

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)](https://ui.adsabs.harvard.edu/abs/1983ApJ...266..713O):
$$
M\_\mathrm{AB} = -2.5\log f\_\nu \,\,–\,\, 48.60.
$$
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](https://docs.astropy.org/en/stable/units/logarithmic_units.html).

> **Comments:**
> **username_3**: I like this explanation.  
> **username_1**: and so do I :-)  

---

