<p>The bolometric correction is the difference between a bolometric magnitude and the magnitude in some band.</p>
<p><span class="math-container">$$BC = M_{\rm bol} - M_{5100} = -2.5\log\left(\frac{L_{\rm bol}}{L_{5100}}\right)
$$</span>
<span class="math-container">$$\log L_{\rm bol} = \log L_{5100} - 0.4BC \ ,$$</span>
<span class="math-container">$$ \log L_{\rm bol} = \log L_{5100} +0.4\log L_{5100} - 21.2  \ ,$$</span>
<span class="math-container">$$ \log L_{\rm bol} = 1.4(1.4 \log L_x -16.8) - 21.2\ , $$</span>
<span class="math-container">$$ \log L_{\rm bol} = 1.96\log L_x  -44.72\ .$$</span></p>

CC BY-SA 4.0
2021-04-02T17:16:43.380Z
2021-04-02T17:16:43.380Z
ProfRob
<p>I am a bit confused by bolometric corrections. If I have an x-ray luminosity in the 2-10 keV band, how does one convert that to <span class="math-container">$L_{bol}$</span>?
From Netzer's book <em>The Physics and Evolution of Active Galactic Nuclei</em> I got these bolometric correction factors:</p>
<p>Optical: <span class="math-container">$BC_{5100} = 53 - log(L_{5100})$</span></p>
<p>and x-ray: <span class="math-container">$log(L_{5100}) = 1.4\times log(L_X) - 16.8$</span></p>
<p>where the bolometric correction for the x-ray luminosity (<span class="math-container">$L_X$</span>) is obtained in two steps, using the equation for the optical BC again. The index <span class="math-container">$5100$</span> stands for the optical continuum measured at <span class="math-container">$5100$</span> angstrom.
I can't figure out what I have to do with this <span class="math-container">$BC_{5100}$</span> once I've got it. Multiply by <span class="math-container">$L_X$</span>? The book says <em>&quot;(...) BCs, that can be used to convert a single-band measurement of <span class="math-container">$L$</span> into an approximate <span class="math-container">$L_{bol}$</span>.&quot;</em></p>
<p>I'm happy to use correction factors defined elsewhere instead of the ones I quoted. I just want to calculate an estimate for <span class="math-container">$L_{bol}$</span> for my galaxies.</p>

<luminosity><x-ray>
How to calculate galaxy bolometric luminosity?
theWrongAlice
# How to calculate galaxy bolometric luminosity?

(Asked by: username_0 on 2021-04-02)

I am a bit confused by bolometric corrections. If I have an x-ray luminosity in the 2-10 keV band, how does one convert that to $L\_{bol}$?
From Netzer's book *The Physics and Evolution of Active Galactic Nuclei* I got these bolometric correction factors:

Optical: $BC\_{5100} = 53 - log(L\_{5100})$

and x-ray: $log(L\_{5100}) = 1.4\times log(L\_X) - 16.8$

where the bolometric correction for the x-ray luminosity ($L\_X$) is obtained in two steps, using the equation for the optical BC again. The index $5100$ stands for the optical continuum measured at $5100$ angstrom.
I can't figure out what I have to do with this $BC\_{5100}$ once I've got it. Multiply by $L\_X$? The book says *"(...) BCs, that can be used to convert a single-band measurement of $L$ into an approximate $L\_{bol}$."*

I'm happy to use correction factors defined elsewhere instead of the ones I quoted. I just want to calculate an estimate for $L\_{bol}$ for my galaxies.

**Tags:** luminosity, x-ray

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## 1 answers

### Answer by username_1 on 2021-04-02. Score: 1 | Accepted answer

The bolometric correction is the difference between a bolometric magnitude and the magnitude in some band.

$$BC = M\_{\rm bol} - M\_{5100} = -2.5\log\left(\frac{L\_{\rm bol}}{L\_{5100}}\right)
$$
$$\log L\_{\rm bol} = \log L\_{5100} - 0.4BC \ ,$$
$$ \log L\_{\rm bol} = \log L\_{5100} +0.4\log L\_{5100} - 21.2 \ ,$$
$$ \log L\_{\rm bol} = 1.4(1.4 \log L\_x -16.8) - 21.2\ , $$
$$ \log L\_{\rm bol} = 1.96\log L\_x -44.72\ .$$


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