How can I calculate the (non-bolometric) luminosity $L$ of a galaxy (or a star for that matter) over a given band from its AB apparent magnitude $m_{AB}$ over that band and its luminosity distance $d_L$?

For instance, consider the g-band which typically has a $\lambda_{eff} = 467 \text{ nm}$ and a $\Delta \lambda = 100 \text{ nm}$. Given this galaxy has an apparent AB magnitude of $m_g = 22.5$ and luminosity distance of $1991 \text{ Mpc}$ (i.e. $z = 0.355$ if you are curious), what is its luminosity? I know I shold use the following equation, but I don't know what value to pick for $M_\odot$.

$$L/L_\odot = 10^{0.4(M_\odot - M)}$$

I tried 5.12 from a website which is the value of $M_\odot$ in g-band but it gives me a different answer compared to when I calculate luminosity using flux:

$$L = 4 \pi d_{L}^2 f_\nu \Delta \nu$$

where $f_\nu$ is flux density and $\Delta \nu$ is the frequency width of the band (this is an approximation to the integration over frequency, assuming bands are step functions). So, how can I find luminosity using absolute magnitude?

Please provide your references.

  • $\begingroup$ How different were your two answers? In addition, 100nm is a pretty wide band to assume uniform flux density. $\endgroup$ Commented Feb 16, 2018 at 13:13
  • $\begingroup$ You are confused. $L$ is usually reserved for a luminosity integrated over all wavelengths. This cannot be calculated simply from the g-band magnitude and distance without knowing more about the galaxy. $\endgroup$
    – ProfRob
    Commented Feb 21, 2018 at 7:06
  • $\begingroup$ @CarlWitthoft, I get magnitude wrong by 1. $\endgroup$
    – Miladiouss
    Commented Feb 21, 2018 at 18:45
  • $\begingroup$ @RobJeffries, In the above question L is the luminosity over the g-band. And it doesn't matter if the flux is not constant because a telescope measures the average flux over a band. So, in this question, L is the luminosity over the g-band and it is not equal to total luminosity and I have made my notation clear by stating "non-bolometric luminosity". Feel free to edit it to $L_{g-band}$. $\endgroup$
    – Miladiouss
    Commented Feb 21, 2018 at 18:45
  • $\begingroup$ Then what are $L_{\odot}$ and $M_{\odot}$ supposed to be? These have very clear defined meanings, and they are bolometric quantities. You also need to tell us what flux (density) and $\Delta \nu$ you used if we are meant to be looking for errors. $\endgroup$
    – ProfRob
    Commented Feb 21, 2018 at 18:52

1 Answer 1


Here is my attempt to reconcile your calculations.

If the AB g-band apparent magnitude is 22.5, then the flux density in the g-band is given by $$f_{\nu} = 10^{(-48.6-22.5)/2.5} = 3.63 \times 10^{-29}\ {\rm erg\ cm}^{-2} {\rm s}^{-1} {\rm Hz}^{-1}$$

If the distance is 1991 Mpc, then the absolute g magnitude is $$ M_g = m_g - 5\log d + 5 = -19.0$$

The absolute AB magnitude of the Sun in the g-band (from your source) is 5.12.

The latter tells us that the ratio of the "g-band luminosites" of the galaxy and the Sun are given by $$\frac{L_g}{L_{\odot,g}} = 10^{0.4(M_{{\odot},g} - M_g)} = 4.45\times 10^{9}$$

One cannot say more than this, in particular one cannot calculate the luminosity of the galaxy, without knowing more about its spectrum.

Also note that the equation above cannot be used to find the ratio of flux in one band to bolometric flux, as I think you are trying to do. To see this, consider that the absolute V-band magnitude and bolometric magnitude of the Sun are almost the same. This does not mean that all the flux from the Sun emerges in the V band!

Your second method requires a figure for $\Delta \nu$, but you haven't said what you have used. The g-filter has a width of around $\Delta \lambda =100$ nm. Using $\lambda \nu = c$ we can say $$ \Delta \nu = |c\Delta \lambda/\lambda^2| = 1.28\times 10^{14}\ {\rm Hz}$$ $$ L_g = 4\pi d^2 f_{\nu} \Delta \nu = 2.2 \times 10^{42} {\rm erg/s}$$

The two figures would agree precisely if the "g-band luminosity" of the Sun were $4.9 \times 10^{32}$ erg/s, or in other words if 13% of the solar luminosity emerged in the g-band as defined above. This does not sound unreasonable, but requires a detailed integration of the solar spectrum over the actual g-band filter profile.

Additional: Just to check the numbers above, we can estimate the flux density of the Sun (in wavelength units, above the atmosphere) to be about 1.7 W m$^{-2}$ nm$^{-1}$ at 467 nm (just search on 'solar spectrum' for many examples) at a distance of $1.5 \times 10^{11}$ m. Putting these numbers together, I calculate a "g-band luminosity" for the Sun of $$ L_{\odot,g}=4\pi d^2 f_{\lambda}\Delta \lambda = 4.8\times 10^{32}\ {\rm erg/s} \ ,$$ which is indeed 13% of the solar bolometric luminosity, and thus $$\frac{L_g}{L_{\odot}} = 0.13\times 4.45\times 10^{9} = 5.79\times 10^8\ .$$

  • $\begingroup$ 13% is too much. This is supposed to be exact since it's almost doing the same calculation with two different methods. I suspect that $L_g$ via flux is correct given we use luminosity distance (distance adjusted for redshift), but further adjustments are needed to use magnitude and we might not be fine using $d_L$ for both methods. I guess what I'm saying is the problem might be due to redshift, but I have to think about it more. $\endgroup$
    – Miladiouss
    Commented Feb 25, 2018 at 4:13
  • $\begingroup$ Also, people don't do $L_g/L_{\odot,g}$, they just do $L_g/L_{\odot}$ which is just an assumed constant of $3.848 \times 10^26$ W. The point is unit having simple units not comparison necessarily. $\endgroup$
    – Miladiouss
    Commented Feb 25, 2018 at 4:16
  • $\begingroup$ @Miladiouss on what basis do say that 13% is too much? The "flux" method is not exact, since you apparently don't know the galaxy spectrum and assuming a top hat filter function is incorrect. $\endgroup$
    – ProfRob
    Commented Feb 25, 2018 at 6:21

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