<p>The extinction <span class="math-container">$A$</span> of a medium (e.g. dust) is defined as the number of magnitudes an object is <em>increased</em> (because higher magnitude means less light), as the light from that object travels through the medium. It is usually (but not always) a function of wavelength <span class="math-container">$\lambda$</span>, and is thus written <span class="math-container">$A_\lambda$</span> (less often <span class="math-container">$A(\lambda)$</span>).</p>
<p>The wavelength can be a single wavelength such as the Lyman <span class="math-container">$\alpha$</span> line at <span class="math-container">$\lambda=1216\,\mathrm{Å}$</span>, in which case we may write that exact wavelength as a subscript, <span class="math-container">$A_{1216}$</span>. Alternatively it can be the total, average extinction in a wavelength band such as the <span class="math-container">$K$</span> band, i.e. the near-infrared wavelength region roughly between <span class="math-container">$\lambda = 2\,\mu\mathrm{m}$</span> and <span class="math-container">$\lambda = 2.5\,\mu\mathrm{m}$</span> (depending on the manufacturer of the filter).</p>
<p>In this case we write <span class="math-container">$A_K$</span>.</p>
<p>I can't seem to find the paper you're referring to, but you ask &quot;How can I get that?&quot;, and the answer is that you measure it. To measure an extinction, you must know how much light you'd <em>expect</em> in the case that some object would <em>not</em> have been obscured (it doesn't necessarily have to be at the particular wavelength you're interested in; it could also just be at a set of other wavelengths, and then you fit an extinction law to these points, under certain assumptions of the properties of the medium). Or you could simply not measure anything, and the assume that the extinction is &quot;similar&quot; to some &quot;standard&quot; extinction law, such as the <a href="https://ui.adsabs.harvard.edu/abs/1989ApJ...345..245C" rel="nofollow noreferrer">Cardelli (1989)</a> law, or the <a href="https://ui.adsabs.harvard.edu/abs/2000ApJ...533..682C" rel="nofollow noreferrer">Calzetti (2000)</a> law.</p>
<h3>Relative extinction</h3>
<p>Because extinction depends not only on the dust properties, but also one the distance traveled, it makes sense to measured it relative to a fixed wavelength, e.g. the optical band <span class="math-container">$V$</span> centered on <span class="math-container">$\lambda \sim 5500\,\mathrm{Å}$</span>, thus factoring out distance. Depending on available data, sometimes the <span class="math-container">$B$</span> band at <span class="math-container">$\lambda \sim 4400\,\mathrm{Å}$</span> is used as reference, but since <span class="math-container">$A_B$</span> is usually larger than <span class="math-container">$A_V$</span>, the flux is larger in <span class="math-container">$V$</span> and hence more precise. From the definition of the color excess
<span class="math-container">$$
E(B-V) \equiv A_B - A_V
$$</span>
and the absolute-to-selective extinction
<span class="math-container">$$
R_V \equiv \frac{A_V}{E(B-V)},
$$</span>
the relation between the two ratios is
<span class="math-container">$$
\frac{A_\lambda}{A_B} = \left(\!\frac{R_V}{1+R_V}\!\right) \frac{A_\lambda}{A_V}
$$</span></p>
<p>You could call the function <span class="math-container">$A_\lambda$</span> the extinction law, but I think it's more common to use the term for the quantity
<span class="math-container">$$
k(\lambda) = \frac{A_\lambda}{E(B-V)}.
$$</span></p>

CC BY-SA 4.0
2021-05-24T23:50:26.680Z
Last question tho. Can I also do/calculate an "absolute-to-selective extinction" for the K-filter or is it only applicable to the optical bandpass?
CGHA
CC BY-SA 4.0
2021-05-24T23:55:25.093Z
oh wait you already mentioned that I can call a function  Aλ. my bad. Thanks.
CGHA
CC BY-SA 4.0
2021-05-24T23:43:31.853Z
2021-05-24T23:43:31.853Z
pela
<p>So I'm calculating V_Alf Ori's Luminosity (or Betelgeuse) using Davies et al. (2013) SED fitted formula from Near-Infrared/NIR K-Band magnitudes and it requires the subtraction of 'extinction'. However, the conventional prescriptions for extinction in the NIR K-regime is a little way off for a massive star with a very dusty environment (likely thick as well due to stellar mass loss and winds enriching it). So instead of using those, I may need to do a quick approximation of it.</p>
<p>This maybe a rookie stuff but as I scour in, which needs some quick clarification or validation, I kept seeing this &quot;<span class="math-container">$A_K$</span>/<span class="math-container">$A_V$</span>&quot; which I presume came from this one?</p>
<p><span class="math-container">$\frac{A_λ}{A_V}$</span> = <span class="math-container">$\frac{R_λE_{(λ-V)}}{R_VE_{(B-V)}}$</span></p>
<p>where;</p>
<ul>
<li><span class="math-container">$R_V$</span> is the canonical Cardelli redenning value of 3.1.</li>
<li><span class="math-container">$A_V$</span> is optical extinction.</li>
<li>E(B-V) is conventional color excess.</li>
<li>While the remaining upper variables are 'almost' the same depending on the bandpass of choice λ.</li>
</ul>
<p>Edit (1): Is that 'correct'/applicable or there's something else on that like applying power law etc.?</p>
<p>Edit (2): Already got the idea of <span class="math-container">$\frac{A_λ}{A_V}$</span>. Then how about <span class="math-container">$A_K$</span>?how can I get that?</p>

CC BY-SA 4.0
2021-05-23T19:20:33.017Z
Can you clarify, what exactly is the question? Are you asking what $A_K$ means, or what the ratio $A_lambda/A_V$ means, or something else?
pela
CC BY-SA 4.0
2021-05-23T21:05:44.963Z
what the $Alambda$/$A_V$ totally means.
CGHA
CC BY-SA 4.0
2021-05-24T10:29:03.683Z
Wait I already got the idea of $A_λ/A_V$ my bad. Well, what would $A_K$ now means?
CGHA
<photometry><extinction>
Clarification on calculating IR Dust Extinction $A_K$/$A_V$
CGHA
# Clarification on calculating IR Dust Extinction $A_K$/$A_V$

(Asked by: username_0 on 2021-05-23)

So I'm calculating V\_Alf Ori's Luminosity (or Betelgeuse) using Davies et al. (2013) SED fitted formula from Near-Infrared/NIR K-Band magnitudes and it requires the subtraction of 'extinction'. However, the conventional prescriptions for extinction in the NIR K-regime is a little way off for a massive star with a very dusty environment (likely thick as well due to stellar mass loss and winds enriching it). So instead of using those, I may need to do a quick approximation of it.

This maybe a rookie stuff but as I scour in, which needs some quick clarification or validation, I kept seeing this "$A\_K$/$A\_V$" which I presume came from this one?

$\frac{A\_λ}{A\_V}$ = $\frac{R\_λE\_{(λ-V)}}{R\_VE\_{(B-V)}}$

where;

* $R\_V$ is the canonical Cardelli redenning value of 3.1.
* $A\_V$ is optical extinction.
* E(B-V) is conventional color excess.
* While the remaining upper variables are 'almost' the same depending on the bandpass of choice λ.

Edit (1): Is that 'correct'/applicable or there's something else on that like applying power law etc.?

Edit (2): Already got the idea of $\frac{A\_λ}{A\_V}$. Then how about $A\_K$?how can I get that?

**Tags:** photometry, extinction

> **Comments:**
> **username_1**: Can you clarify, what exactly is the question? Are you asking what $A_K$ means, or what the ratio $A_lambda/A_V$ means, or something else?  
> **username_0**: what the $Alambda$/$A_V$ totally means.  
> **username_0**: Wait I already got the idea of $A_λ/A_V$ my bad. Well, what would $A_K$ now means?  

---

## 1 answers

### Answer by username_1 on 2021-05-24. Score: 3 | Accepted answer

The extinction $A$ of a medium (e.g. dust) is defined as the number of magnitudes an object is *increased* (because higher magnitude means less light), as the light from that object travels through the medium. It is usually (but not always) a function of wavelength $\lambda$, and is thus written $A\_\lambda$ (less often $A(\lambda)$).

The wavelength can be a single wavelength such as the Lyman $\alpha$ line at $\lambda=1216\,\mathrm{Å}$, in which case we may write that exact wavelength as a subscript, $A\_{1216}$. Alternatively it can be the total, average extinction in a wavelength band such as the $K$ band, i.e. the near-infrared wavelength region roughly between $\lambda = 2\,\mu\mathrm{m}$ and $\lambda = 2.5\,\mu\mathrm{m}$ (depending on the manufacturer of the filter).

In this case we write $A\_K$.

I can't seem to find the paper you're referring to, but you ask "How can I get that?", and the answer is that you measure it. To measure an extinction, you must know how much light you'd *expect* in the case that some object would *not* have been obscured (it doesn't necessarily have to be at the particular wavelength you're interested in; it could also just be at a set of other wavelengths, and then you fit an extinction law to these points, under certain assumptions of the properties of the medium). Or you could simply not measure anything, and the assume that the extinction is "similar" to some "standard" extinction law, such as the [Cardelli (1989)](https://ui.adsabs.harvard.edu/abs/1989ApJ...345..245C) law, or the [Calzetti (2000)](https://ui.adsabs.harvard.edu/abs/2000ApJ...533..682C) law.

### Relative extinction

Because extinction depends not only on the dust properties, but also one the distance traveled, it makes sense to measured it relative to a fixed wavelength, e.g. the optical band $V$ centered on $\lambda \sim 5500\,\mathrm{Å}$, thus factoring out distance. Depending on available data, sometimes the $B$ band at $\lambda \sim 4400\,\mathrm{Å}$ is used as reference, but since $A\_B$ is usually larger than $A\_V$, the flux is larger in $V$ and hence more precise. From the definition of the color excess
$$
E(B-V) \equiv A\_B - A\_V
$$
and the absolute-to-selective extinction
$$
R\_V \equiv \frac{A\_V}{E(B-V)},
$$
the relation between the two ratios is
$$
\frac{A\_\lambda}{A\_B} = \left(\!\frac{R\_V}{1+R\_V}\!\right) \frac{A\_\lambda}{A\_V}
$$

You could call the function $A\_\lambda$ the extinction law, but I think it's more common to use the term for the quantity
$$
k(\lambda) = \frac{A\_\lambda}{E(B-V)}.
$$

> **Comments:**
> **username_0**: Last question tho. Can I also do/calculate an "absolute-to-selective extinction" for the K-filter or is it only applicable to the optical bandpass?  
> **username_0**: oh wait you already mentioned that I can call a function  Aλ. my bad. Thanks.  

---

