<p>First of all, your first question.</p>

<p><a href="http://www.iisc.ernet.in/currsci/apr102004/921.pdf" rel="nofollow">This source</a> clearly state that <code>Values are given in the usual logarithmic (dex) scale</code>, 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.</p>

<p>I think it is better to work out with an example. Let's take the <code>He</code>. </p>

<p>Good enough, you can better read the paper from <a href="http://astro.uni-tuebingen.de/~rauch/TMAP/grevessesauval2001.pdf" rel="nofollow">here</a>.</p>

<p>From the table, we have $A_{el}=10.93$. This is <em>the <strong>abundance</strong> of He relative to H</em> (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).</p>

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

CC BY-SA 3.0
2014-10-24T16:18:21.863Z
So, not natural logarithm.
NeutronStar
CC BY-SA 3.0
2014-10-24T16:28:22.920Z
That reference says (dex), and you can obtain 8% only by decimal logarithm.
Py-ser
CC BY-SA 3.0
2014-10-24T03:32:25.677Z
2014-10-24T03:32:25.677Z
Py-ser
<p>Close to the bottom of page 4 of <a href="http://link.springer.com/article/10.1023/A:1005161325181#page-1" rel="nofollow">this article</a> (marked as page 164 in the upper left corner) states </p>

<blockquote>
  <p>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.</p>
</blockquote>

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

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

<p>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</p>

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

<p><strong>Is this the correct way to calculate the relative abundances?</strong>  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$_e$.</p>

CC BY-SA 3.0
2015-12-25T00:05:08.940Z
For abundances Log means Log_10.
earnric
CC BY-SA 3.0
2014-10-23T03:10:45.657Z
What does it mean the bottom? Does not this paper have page/equation numbering?
Py-ser
CC BY-SA 3.0
2014-10-23T11:59:14.403Z
@Py-ser, no the equations are not numbered.  It's an inline equation to boot.
NeutronStar
CC BY-SA 3.0
2014-10-24T01:27:18.433Z
I guess the page are numbered yet. The paper is a bit tough, please give the page number.
Py-ser
CC BY-SA 3.0
2014-10-24T02:34:01.660Z
@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)"
NeutronStar
CC BY-SA 3.0
2014-10-24T03:02:40.187Z
Ops, sorry my mistake. I think I have the sources for an answer in a while.
Py-ser
<the-sun><hydrogen><elemental-abundances><helium>
How are stellar elemental abundances quoted?
NeutronStar
# How are stellar elemental abundances quoted?

(Asked by: username_0 on 2014-10-22)

Close to the bottom of page 4 of [this article](http://link.springer.com/article/10.1023/A:1005161325181#page-1) (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$\_e$.

**Tags:** the-sun, hydrogen, elemental-abundances, helium
> **Comments:**
> **username_2**: What does it mean the bottom? Does not this paper have page/equation numbering?  
> **username_0**: @Py-ser, no the equations are not numbered.  It's an inline equation to boot.  
> **username_2**: I guess the page are numbered yet. The paper is a bit tough, please give the page number.  
> **username_0**: @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)"  
> **username_2**: Ops, sorry my mistake. I think I have the sources for an answer in a while.  
> **username_1**: For abundances Log means Log_10.  

---

## 1 answers

### Answer by username_2 on 2014-10-24. Score: 2 | Accepted answer

First of all, your first question.

[This source](http://www.iisc.ernet.in/currsci/apr102004/921.pdf) 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](http://astro.uni-tuebingen.de/~rauch/TMAP/grevessesauval2001.pdf).

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 $25\%$.

> **Comments:**
> **username_0**: So, not natural logarithm.  
> **username_2**: That reference says (dex), and you can obtain 8% only by decimal logarithm.  

---

