How to calculate galaxy bolometric luminosity?

Question

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.

Answer 1

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

$$BC = M_{\rm bol} - M_{5100} = -2.5\log\left(\frac{L_{\rm bol}}{L_{5100}}\right) $$ $$\log L_{\rm bol} = \log L_{5100} - 0.4BC \ ,$$ $$ \log L_{\rm bol} = \log L_{5100} +0.4\log L_{5100} - 21.2 \ ,$$ $$ \log L_{\rm bol} = 1.4(1.4 \log L_x -16.8) - 21.2\ , $$ $$ \log L_{\rm bol} = 1.96\log L_x -44.72\ .$$