I surfed the internet and searched a lot of sites, so far I got this

enter image description here

but unable to find the formula which relates only Muv and the Ψ as I don't know the value of parameters Auv and μ(z).

If there exists any such formula, then do help!

  • $\begingroup$ Explain all the terms please. $\endgroup$
    – ProfRob
    Nov 12, 2020 at 20:27
  • $\begingroup$ Hi space_nkc, I don’t know if you saw it, but I edited my answer after you accepted it, to give a simpler relation :) $\endgroup$
    – pela
    Nov 14, 2020 at 23:52

1 Answer 1


The UV luminosity of a galaxy can be calculated, given a stellar population. This population, in turn, can be calculated given an initial mass function (IMF), i.e. the distribution function of stellar masses. In this case, the UV luminosity should be linearly proportional to the star formation rate (SFR), sometimes written $\Psi$.

The UV is primarily emitted by young stars with lifetimes of $\lesssim10^8\,\mathrm{yr}$. Assuming that the SFR is constant over longer time scales, Kennicutt (1998) calculates in his seminal paper (with ~5000 citations) for a Salpeter (1995) IMF (equally seminal; ~6700 citations) with mass limits 0.1 and 100 $M_\odot\,\mathrm{yr}^{-1}$ that $$ \frac{\Psi_\mathrm{Sal}}{M_\odot\,\mathrm{yr}^{-1}} = 1.4\times10^{-28}\frac{L_\nu}{\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{Hz}^{-1}}. $$ were $L_\nu$ is the emitted UV luminosity density, i.e. luminosity per frequency bin. At a (luminosity) distance $d_L$ from the galaxy, the observed flux density $f_\nu$ is then $$ f_\nu = \frac{L_\nu}{4\pi d_L^2}. $$

The flux density is related to the apparent (observed) UV magnitude $m_\mathrm{UV}$ as (Oke & Gunn 1983; ~1900 citations despite their +/- error in their most important equation) $$ m_\mathrm{UV} = -2.5\log \frac{f_\nu}{\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-1}\,\mathrm{Hz}^{-1}} - 48.6. $$

Distance modulus and dust extinction

The apparent magnitude is related to the absolute magnitude $M$ through the distance modulus $$ \mu \equiv m - M. $$

Furthermore, the above relations don't take into account the fact that some of the emitted light is absorbed by dust. Dust extinction increases the observed magnitude (because magnitudes are backwards) by a number $A_\mathrm{UV}$.

That is, the apparent UV magnitude is $$ m_\mathrm{UV} = M_\mathrm{UV} + \mu + A_\mathrm{UV}. $$

Typically, the extinction $A_\mathrm{UV}$ isn't measured explicitly (because UV measurements are difficult); rather the extinction is measured in two optical bands, e.g. $B$ and $V$. The extinction $A_\lambda$ at a given wavelength $\lambda$ can then be estimated as $$ A_\lambda = k_\lambda E(B-V), $$ where $E(B-V)\equiv A_B - A_V$ is the color excess, and $k_\lambda$ is given by the assumed extinction law, e.g. a Cardelli (1989) law or a Calzetti (2000) law.

The role of the initial mass function

The Salpeter IMF is old, and at that time we didn't have too good data for the low-mass stars (which are faint), so he just fitted a single power-law. With better data, it was eventually found that there were less low-mass stars than previously thought, i.e. it takes less stellar mass to produce a given UV luminosity. A popular choice today is the Chabrier (2003) IMF (~5100 citations), which gives roughly a factor of 1.8 more UV than the Salpeter IMF per star formed, on average. That is, the conversion factor is 1.8 times smaller.

With a Chabrier IMF, you hence get that the relation between absolute magnitude of UV and star formation rate (which is what you're asking) is $$ \frac{\Psi_\mathrm{Cha}}{M_\odot\,\mathrm{yr}^{-1}} = 7.8\times10^{-29}\,\times\,4\pi d_L^2 \times10^{-(M_\mathrm{UV} + \mu + A_\mathrm{UV}+48.6)/2.5}, $$ where $d_L$ is measured in cm.

Simpler relation

The above formula resembles the one you give, but is a bit strange in that it both contains $\mu$ and $d_L$, which are just two different measures of the same thing. Using the definition $$ \mu = 5\log\left(\frac{d_L}{10\,\mathrm{pc}}\right) $$ we can — after a bit of algebra — express the definition of the Oke & Gunn (AB) magnitude as $$ M_\mathrm{UV} = -2.5\log L_\nu + 51.6. $$ With this, we can can write the relation as $$ \frac{\Psi_\mathrm{Cha}}{M_\odot\,\mathrm{yr}^{-1}} = 7.8\times10^{-29} \,\times 10^{-(M_\mathrm{UV}+A_\mathrm{UV}-51.6)/2.5}, $$ or even $$ \boxed{ \frac{\Psi_\mathrm{Cha}}{M_\odot\,\mathrm{yr}^{-1}} = 3.4\times10^{-8} \,\times 10^{-(M_\mathrm{UV}+A_\mathrm{UV})/2.5}.} $$

  • $\begingroup$ @RobJeffries Hmm… yeah, maybe… But the title asks for the relation between absolute, not apparent, magnitude and SFR, so the way I interpreted this is that they found that formula, but didn't know what the terms meant. But I agree it's a somewhat peculiar formula, so I'll edit to give a simpler form. $\endgroup$
    – pela
    Nov 13, 2020 at 21:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .