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I have the dust-corrected luminosity in the Ks band in the unit erg/s/Hz. I want to convert it to erg/s. I would like to know which method is correct:

  1. $L[erg/s]=L[erg/s/Hz]\times(C/\lambda)$
  2. $L[erg/s]=L[erg/s/Hz]\times(C/\lambda^2)d\lambda$

For the Ks band, the central wavelength is 2154 nm and the width is 309 nm. For example, with the first method, for Id=219816 from Cosmos 2015 and log[L]=28.324 [erg/s/Hz], the answer is

$7.67\times {10}^{8} L_\odot$ and with the second method, it is

$1.1\times {10}^{8} L_\odot$

I don’t know which one is correct. My goal is to multiply this by the mass-to-light ratio of simulation data to determine the mass. I should mention that the redshift of this source is 1.104.

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  • $\begingroup$ What do you mean by "luminosity"? If you mean the $L$ in the mass-to-light ration then this cannot be straightforwardly estimated from the spectral flux in a single band. The second method gives (approximately) the energy per second emerging in the K-band. $\endgroup$
    – ProfRob
    Commented Apr 13 at 11:27
  • $\begingroup$ (Do cosmologists actually use such a unit as erg/s/Hz? Doesn't /s/Hz just cancel out?) $\endgroup$
    – Karl Knechtel
    Commented Apr 13 at 11:45
  • $\begingroup$ Actually, the mass-to-light ratio is denoted as $M/L_{K_s}$ , and this ratio is specific to a certain band. I don’t understand why you say we cannot use this specific band. When I use the absolute magnitude ($M_{K_s}=-18.585$ in the AB system), I get this result: $2.77\times {10}^{9} L_\odot$. This is ten times larger than the result obtained from the second method. $\endgroup$
    – Fatemeh Abedini
    Commented Apr 13 at 12:03
  • $\begingroup$ What is the mass to light ratio of the Sun in that system ($M/L_K$)? $\endgroup$
    – ProfRob
    Commented Apr 13 at 18:00
  • $\begingroup$ @KarlKnechtel Yes, definitely, although formally it's called luminosity density. The spectrum of a luminous source is a function of wavelength or, equivalently, frequency, so is given a emitted energy (erg), per second (s), per frequency bin (Hz) or wavelength bin (Å). Although you could let the s and the Hz cancel, it's more explanatory to leave it. If you integrate the luminosity density over some frequency (or wavelength) interval, then you get the total luminosity in that interval. $\endgroup$
    – pela
    Commented Apr 13 at 18:25

1 Answer 1

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To make it easier to discuss, consider the terminology listed here. Since the first value has units of [erg/s/Hz], this is a spectral flux (F$_{\nu}$). Luminosity (L) has units of [erg/s]. By definition, spectral flux is the luminosity per unit Hz. So to get the luminosity, you would generally integrate the flux over the observed band:

$L=\int F_{\nu}(\nu) d\nu$

But in your case, you may simply multiply by the width of the band (method 2). Because you have a wavelength width, you may use the fact that $\nu=c/\lambda$ to derive the relevant width: $\Delta \nu = \frac{c\Delta\lambda}{\lambda^2}$. This results in the form:

$L=F_{\nu}(\nu) \Delta\nu=F_{\nu}(\nu)\frac{c\Delta\lambda}{\lambda^2}$

Or, using your units:

$L[erg/s]=F_{\nu}(\nu)[erg/s/Hz]\,c[m/s]\,\Delta\lambda[m]\,(\lambda[m])^{-2}$

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  • $\begingroup$ Thanks a lot, works. $\endgroup$
    – Fatemeh Abedini
    Commented Jul 18 at 7:58

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