<p>To make it easier to discuss, consider the terminology listed <a href="https://en.wikipedia.org/wiki/Flux#SI_radiometry_units" rel="nofollow noreferrer">here</a>. Since the first value has units of [erg/s/Hz], this is a spectral flux (F<span class="math-container">$_{\nu}$</span>). 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:</p>
<p><span class="math-container">$L=\int F_{\nu}(\nu) d\nu$</span></p>
<p>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 <span class="math-container">$\nu=c/\lambda$</span> to derive the relevant width: <span class="math-container">$\Delta \nu = \frac{c\Delta\lambda}{\lambda^2}$</span>. This results in the form:</p>
<p><span class="math-container">$L=F_{\nu}(\nu) \Delta\nu=F_{\nu}(\nu)\frac{c\Delta\lambda}{\lambda^2}$</span></p>
<p>Or, using your units:</p>
<p><span class="math-container">$L[erg/s]=F_{\nu}(\nu)[erg/s/Hz]\,c[m/s]\,\Delta\lambda[m]\,(\lambda[m])^{-2}$</span></p>

CC BY-SA 4.0
2024-07-18T07:58:27.573Z
Thanks a lot, works.
Fatemeh Abedini
CC BY-SA 4.0
2024-07-16T14:26:02.300Z
2024-07-16T14:26:02.300Z
gj283
<p>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:</p>
<ol>
<li><span class="math-container">$L[erg/s]=L[erg/s/Hz]\times(C/\lambda)$</span></li>
<li><span class="math-container">$L[erg/s]=L[erg/s/Hz]\times(C/\lambda^2)d\lambda$</span></li>
</ol>
<p>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</p>
<p><span class="math-container">$7.67\times {10}^{8} L_\odot$</span>
and with the second method, it is</p>
<p><span class="math-container">$1.1\times {10}^{8} L_\odot$</span></p>
<p>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.</p>

CC BY-SA 4.0
2024-04-13T11:27:51.907Z
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.
ProfRob
CC BY-SA 4.0
2024-04-13T11:45:33.107Z
(Do cosmologists actually use such a unit as erg/s/Hz? Doesn't /s/Hz just cancel out?)
Karl Knechtel
CC BY-SA 4.0
2024-04-13T12:03:35.263Z
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.
Fatemeh Abedini
CC BY-SA 4.0
2024-04-13T18:00:09.770Z
What is the mass to light ratio of the Sun in that system ($M/L_K$)?
ProfRob
CC BY-SA 4.0
2024-04-13T18:25:54.463Z
@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.
pela
CC BY-SA 4.0
2024-04-19T10:51:15.697Z
@ProfRob Sorry for the late reply. The $M/{L_{K_s}}=0.17709$, and I should mention that the redshift of this source is 1.104.
Fatemeh Abedini
<observational-astronomy><galaxy><cosmology><spectra><luminosity>
How can I convert luminosity from [erg/s/Hz] to [erg/s]
Fatemeh Abedini
# How can I convert luminosity from [erg/s/Hz] to [erg/s]

(Asked by: username_0 on 2024-04-13)

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.

**Tags:** observational-astronomy, galaxy, cosmology, spectra, luminosity
> **Comments:**
> **username_1**: 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.  
> **username_2**: (Do cosmologists actually use such a unit as erg/s/Hz? Doesn't /s/Hz just cancel out?)  
> **username_0**: 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.  
> **username_1**: What is the mass to light ratio of the Sun in that system ($M/L_K$)?  
> **username_3**: @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.  
> **username_0**: @ProfRob Sorry for the late reply. The $M/{L_{K_s}}=0.17709$, and I should mention that the redshift of this source is 1.104.  

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## 1 answers

### Answer by username_4 on 2024-07-16. Score: 1

To make it easier to discuss, consider the terminology listed [here](https://en.wikipedia.org/wiki/Flux#SI_radiometry_units). 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:

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

> **Comments:**
> **username_0**: Thanks a lot, works.  

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