3
$\begingroup$

Could someone please tell me if the ratio $$\frac{L_B}{L_{H\alpha}}$$

is important in determining star formation rates?

Additionally, could someone please explain the implication of the ratio to me or direct me to a source?

$\endgroup$
2
  • 2
    $\begingroup$ You should give more details like what are the terms in this ratio to make your question clearer. $\endgroup$
    – MBR
    Dec 14 '17 at 8:53
  • 1
    $\begingroup$ Is $L_B$ the $B$ band luminosity? If so, the ratio is a color. That doesn't really constrain the SFR (although larger SFRs generally lead to bluer colors). But your denominator can be used alone, through the Kennicutt (1998) relation: $\mathrm{SFR} = 7.9\times10^{-42} L_{\mathrm{H}\alpha}$. $\endgroup$
    – pela
    Dec 14 '17 at 12:09
1
$\begingroup$

I assume that $L$ stands for lumniosity, i.e. energy emitted per time interval, and the index is refering to the respective band:

  • $L_{H \alpha}$ is the lumniosity of the visible spectral line in the Balmer series with $656.28 {\rm nm}$ wavelength
  • $L_B$ might be the lumniosity for B band, i.e. for radio frequencies between $250\ldots 500 {\rm MHz}$, or for blue light of wavelength $445 {\rm nm}$ with FWHM of $94 {\rm nm}$, as defined by the photometric system - which is the more proable assumption.

There are some star-formation-rate indicators based on lumniosity, but those seem to be based on a single band:

[...] with constant star formation of 100 Myr, the non-ionising UV $(0.0912 \mu{\rm m} < \lambda < 0.3 \mu {\rm m})$ stellar continium can be converted to a SFR:

$$ SFR(UV) = 3.0 \cdot 10^{-47} \lambda \, L(\lambda)$$ with SFR(UV) in $M_\odot {\rm yr}^{-1}$, $\lambda$ in $\overset{\circ}{A}$, and $L(\lambda)$ in erg/s.

As @pela already mentioned in 2017, there is some relationship between star formation rate (SFR) and lumniosity of the ${\rm H \alpha}$ line alone, here cited from Daniela Calzetti's web paper, which essentially is arXiv:1208.2997

Snapshot of above mentioned website

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.