How to work properly with luminosities computed for a given filter?

[liempi]

Hi all.

I'm new working with the luminosities computed by Galacticus, and I think it could be useful to ask here to keep the answer visible.

I computed the luminosities using in a rest (and observed) frame using the JWST_NIRCAM_f090w filter available.
Then Galacticus outputs the stellar luminosities at a given redshift in units of W/Hz according to the physics manual. So, what is the best approach to obtain the luminosity in watts?

Many thanks :)

[abensonca]

A few important points here:

1. The luminosities are not output in units of W/Hz. They're output in units of $4.4659 \times 10^{13}$ W/Hz which is the zero point of the AB magnitude system.
2. Usually, in astronomical observations, you don't want the actual luminosity (e.g. the luminosity in Watts as you are asking about), because luminosities are almost always measured through a filter (often the combination of an actual filter and the detector response etc.). That's why the usual quantity measured is the luminosity relative to a source of constant luminosity density observed through the same filter, i.e.:

$$ \tilde{L} = \left. \int_0^\infty \frac{S(\nu)}{\mathrm{h}\nu} R(\nu) \mathrm{d}\nu \right/ \int_0^\infty \frac{S_\mathrm{AB}}{\mathrm{h}\nu} R(\nu) \mathrm{d}\nu $$

where $S(\nu)$ is the spectral density (W/Hz) of the source, $S_\mathrm{AB}$ is the spectral density for the AB magnitude system, and $R(\nu)$ is the filter response (fraction of photons that pass through the filter. Note that the $\mathrm{h}\nu$ term is included as most (at least most modern) sensors count photons.
3. If you really want a luminosity - a thing measured in W, you need to first think carefully about what you actually want. You could define some luminosity $\hat{L}$:

$$ \hat{L} = \tilde{L} \mathrm{h}\bar{\nu} \int_0^\infty \frac{S_\mathrm{AB}}{\mathrm{h}\nu} R(\nu) \mathrm{d}\nu =\mathrm{h}\bar{\nu} \ \int_0^\infty \frac{S(\nu)}{\mathrm{h}\nu} R(\nu) \mathrm{d}\nu $$

where $\bar{\nu}$ is some mean frequency in the filter. This isn't very meaningful though - it depends on the filter shape (and, this is exactly why we typically don't measure these).

If you really wanted to know the total luminosity in some frequency range you could define a filter that was proportional to $\nu$ in that range and zero outside, the integral would then effectively become:

$$ \tilde{L} = \left. \int_{\nu_\mathrm{min}}^{\nu_\mathrm{max}} S(\nu) \mathrm{d}\nu \right/ \int_{\nu_\mathrm{min}}^{\nu_\mathrm{max}} S_\mathrm{AB} \mathrm{d}\nu = \left. \int_{\nu_\mathrm{min}}^{\nu_\mathrm{max}} S(\nu) \mathrm{d}\nu \right/ (S_\mathrm{AB} [\nu_\mathrm{max} - \nu_\mathrm{min}]) $$

which you could then multiply by $S_\mathrm{AB} [\nu_\mathrm{max} - \nu_\mathrm{min}]$ to get the total luminosity. This is not a quantity you can compared to anything observed though, so I doubt that it's useful.

So, what are you trying to accomplish with the luminosities that you are outputting? If you explain that in more detail I can probably suggest the best solution.

[liempi]

I want to compute the color of a galaxy given two filters and also construct the luminosity function.

[abensonca]

So, probably what Galacticus is outputting is exactly what you want! Colors are typically defined in terms of magnitudes. If the luminosity as output by Galacticus is $L$, then the corresponding absolute magnitude is just $M = - 2.5 \log_{10} L$. The color of the object is then just the difference in the magnitude between the two filters.

For the luminosity function, is there specific data you're trying to compare to?

[liempi]

Many thanks!

For the luminosity function there is no dataset yet, but I plan to compare with the this work, where the AGN bolometric luminosity function is computed for some galaxies which I want to select from a color-magnitude criteria.

[abensonca]

I see. Selecting from a color-magnitude criterion should be straightforward in that case.

[liempi]

Yes, it was unclear to me how to start working with the luminosities, but now the procedure is clear.

Many thanks again!