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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.

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    $\begingroup$ No particular obvious answer to your problem but yes, your M_J=12.5 is way too faint for giants. From this table for dwarfs, expected M_J is between 6 (M0V) and 11.6 (M9V) $\endgroup$ Aug 18, 2023 at 20:02
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    $\begingroup$ Thank you - I suspected it was far too faint. It's frustrating as I'm pretty sure the maths is correct. $\endgroup$
    – Rumi
    Aug 19, 2023 at 8:52
  • $\begingroup$ Interesting question and Welcome to Stack Exchange! I changed $Euclid zp$ into $\text{Euclid zp}$ using $\text{Euclid zp}$ but I am not sure what it is - could you clarify it for those of us unfamiliar with the term? Thanks! $\endgroup$
    – uhoh
    Aug 21, 2023 at 23:03
  • $\begingroup$ Please explain what $F$ and $T$ are. $\endgroup$
    – ProfRob
    Aug 22, 2023 at 6:02
  • $\begingroup$ Euclid zp is the zero point that's given in the Euclid telescope release papers. T is the total response for the system, also released with those papers. F is the flux density from the first bullet point. $\endgroup$
    – Rumi
    Sep 4, 2023 at 8:58

2 Answers 2

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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).

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  • $\begingroup$ Transmission functions are required because filter bandpasses are not top hats. $\endgroup$
    – ProfRob
    Aug 22, 2023 at 6:22
  • $\begingroup$ Yep, I'm trying to get an idea of what the absolute magnitudes will be for different spectral types when Euclid data reaches me, so I'm using the Euclid transmission functions. $\endgroup$
    – Rumi
    Sep 4, 2023 at 9:30
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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.

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  • $\begingroup$ I'm likely overcomplicating things, but I'm working in the AB magnitude system and from what I've read it's all quoted in terms of $f_\nu$ and so the $c$ comes from converting between $f_\nu$ and $f_\lambda$. So I'm starting with a synthetic spectrum, with units [ergs/s], and then dividing by area and wavelength to get to a wavelength flux density, and then I'm plugging it into the equations I've listed in my original question. There are some details in the Pyphot documentation pages. $\endgroup$
    – Rumi
    Sep 4, 2023 at 9:20

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