Close to the bottom of page 4 of this article (marked as page 164 in the upper left corner) states

Values are given in the logarithmic scale usually adopted by astronomers, A$_{e\ell}$ = log N$_{e\ell}$/N$_H$ + 12.0, where N$_{e\ell}$ is the abundance by number.

I have never used stellar elemental abundances in this way, so I want to check a few things.

First of all, is the logarithm used by this paper log$_{10}$ or log$_e$ (often written ln)? My guess is log$_e$ because of what follows.

Let's say I want to calculate the elemental abundance of He relative to H. I believe the correct math would be to do the following

$$\frac{\textrm{N}_{He}}{\textrm{N}_H} = \textrm{exp}(\textrm{A}_{He} - 12.0)$$

Is this the correct way to calculate the relative abundances? This equation gives a value of 0.343 for the helium abundance relative to hydrogen, which seems to be the correct value for the sun (the sun is usually quotes as being about 70% H and about 25% He, which gives a relative abundance of 0.357). This correct value is what leads me to believe that the log is log$_e$.

  • $\begingroup$ What does it mean the bottom? Does not this paper have page/equation numbering? $\endgroup$
    – Py-ser
    Oct 23, 2014 at 3:10
  • $\begingroup$ @Py-ser, no the equations are not numbered. It's an inline equation to boot. $\endgroup$ Oct 23, 2014 at 11:59
  • $\begingroup$ I guess the page are numbered yet. The paper is a bit tough, please give the page number. $\endgroup$
    – Py-ser
    Oct 24, 2014 at 1:27
  • $\begingroup$ @Py-ser I already did in my original post of the question. To quote myself, "Close to the bottom of page 4 of this article (marked as page 164 in the upper left corner)" $\endgroup$ Oct 24, 2014 at 2:34
  • $\begingroup$ Ops, sorry my mistake. I think I have the sources for an answer in a while. $\endgroup$
    – Py-ser
    Oct 24, 2014 at 3:02

1 Answer 1


First of all, your first question.

This source clearly state that Values are given in the usual logarithmic (dex) scale, for the same formula that you quoted (similar job). It is a bit tricky as the article "explains" the values, but you have to pay attention to the exact definition.

I think it is better to work out with an example. Let's take the He.

Good enough, you can better read the paper from here.

From the table, we have $A_{el}=10.93$. This is the abundance of He relative to H (in logarithmic scale). From this you find out that $\frac{N_{el}}{N_H}=0.08 = 8\%$. Indeed the work confirms this value (see the last page).

What you quote as about 25% He, is what they call abundances by mass of [...] Helium (Y), which means $Y = mass\ of\ Helium / mass\ of\ Hydrogen$, and this is indeed about $25\%$.

  • $\begingroup$ So, not natural logarithm. $\endgroup$ Oct 24, 2014 at 16:18
  • $\begingroup$ That reference says (dex), and you can obtain 8% only by decimal logarithm. $\endgroup$
    – Py-ser
    Oct 24, 2014 at 16:28

You must log in to answer this question.

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