How to get the flux from a synthetic spectrum in a bandpass, when only given specific luminosity times frequency?

Question

I have some synthetic spectra of stars (M and K giants) with columns wavelength [Angstroms], continuum normalised flux and specific luminosity $\times$ frequency [ergs/s]; I also have some Euclid filter transmission curves.

I wish to extract the flux within the filter, and then convert it into AB magnitudes. My process for doing this has been:

• convert specific luminosity $\times$ frequency into flux density by dividing by wavelength and area $4 \pi d^2$, assuming the star is at 10pc to end up with the absolute magnitude. After this step it now has units [erg/s/Å/cm²].
• Resample the data to match the wavelength grid of the system response.
• Calculate the flux in the bandpass according to

$$f_\nu = \frac{1}{c}\frac{\int FT\lambda\,d\lambda}{\int T / \lambda\,d\lambda}$$

• Then finally calculate the magnitude as $m = -2.5log(f_\nu) + \text{Euclid zp}$

However, I'm not getting sensible values (absolute mag around 12.5 in Euclid J-band). I've also tried using Pyphot, which gives me roughly the same values, and their source code is very similar to what I originally had. It makes me think there is something wrong with the assumptions I'm making about getting flux density from specific luminosity multiplied by frequency. I'd really appreciate it if anyone could set me on the right path.

Answer 1

You should not be dividing by wavelength in your first step. The values in each given wavelength interval have already the dimension of luminosity ($erg/s$), so you just have to divide these by $4\pi d^2$ to get the flux ($erg/s/cm^2$). Then just add up all the intervals and convert the final result to absolute magnitude (I don't know why you would need filter transmission curves for this if you have synthetic spectra rather than measured ones).

Answer 2

Pretty sure it is step 3. I can't understand your integrals. If $F$ is the spectral flux and $T$ is the transmission function at each wavelength, then the average flux through the filter is just $$ f = \frac{\int FT\ d\lambda}{\int T\ d\lambda}\ . $$ This has units of erg/s/cm$^2$/A.

I have no idea why $c$ features unless there is something about the units you haven't told us. It is possible you are trying to convert to an average flux $f_\nu$ in frequency units because that is how the zeropoint is defined? In which case, you can convert $F$ into $F_\nu$ $$F_\nu = F \left|\frac{d\lambda}{d\nu}\right| = \frac{F\lambda^2}{c}\ . $$

Then: $$f_\nu = \frac{\int F_\nu T\ d\nu}{\int T\ d\nu} $$ as long as you have the transmission defined as a function of frequency.

More information is required in the question.