<ul>
<li><span class="math-container">$\gamma$</span> = emission angle, the direction from a point on the surface of the limb-darkened body (presumably a star) toward the observer, with respect to the local vertical direction (0° at center of disk, 90° at edge of disk)</li>
<li><span class="math-container">$\mu$</span> = cos(<span class="math-container">$\gamma$</span>) = the emission angle cosine</li>
</ul>
<p>If I infer that you're asking about the limb darkening of a star that is being transited by a planet, then neither the star nor the planet is resolved. But as the planet moves across the star, it will block the same angular area at different parts of the transit, but the brightness of the spot it's blocking will vary because of limb darkening. So this causes the <a href="https://www.hao.ucar.edu/research/stare/hd209458.html" rel="nofollow noreferrer">transit lightcurve to have a non-flat bottom</a>.</p>
<p>A transit seen perfectly edge-on will have the planet blocking <span class="math-container">$\gamma$</span> = 0° at the midpoint of the transit, like you say in the original question. But usually transits are seen at some finite inclination, and so the planet blocks <span class="math-container">$\gamma$</span> = 90° at the start of the transit, but only reaches <span class="math-container">$\gamma_{\rm{max}}$</span> at the midpoint, where 0° &lt; <span class="math-container">$\gamma_{\rm{max}}$</span> &lt; 90°. The value of <span class="math-container">$\gamma_{\rm{max}}$</span> will depend on the orbital radius, stellar diameter, and viewing inclination angle.</p>

CC BY-SA 4.0
2022-01-17T18:59:11.583Z
2022-01-17T18:59:11.583Z
giardia
<p>To calculate <span class="math-container">$\mu$</span> of a point on the Sun:
<span class="math-container">$\text{azimuth of point} = \arcsin(\frac{d}{r}),
\mu = \cos(\text{azimuth of point})$</span></p>
<p>where d=distance of point from centre of disc, and r is the radius of the disk.</p>
<p>You can try visualising it from the side of the disc. Here's a nice figure (though I would exchange the r and a notation for more clarity, as r usually denotes radius): <a href="https://phys.libretexts.org/Bookshelves/Astronomy__Cosmology/Book%3A_Stellar_Atmospheres_(Tatum)/06%3A_Limb_Darkening/6.01%3A_Introduction._The_Empirical_Limb-darkening" rel="nofollow noreferrer">https://phys.libretexts.org/Bookshelves/Astronomy__Cosmology/Book%3A_Stellar_Atmospheres_(Tatum)/06%3A_Limb_Darkening/6.01%3A_Introduction._The_Empirical_Limb-darkening</a></p>
<p>Mu will be 1 if the point is in disc centre (i.e., azimuth is 0 deg.), and it will be 0, if it is on the limb (i.e., azimuth is 90 deg.). So what it tells you basically is how far the point is from disc centre. That's why it is used to describe limb darkening.</p>
<p>I am not from the community that looks at planet transits, but I guess one could describe that also with mu, and then if it is right in the centre of the observed disc, azimuth would be indeed 0 degrees. However, as I understand, it should be referring to something on the surface, rather than something passing in front of the disc.</p>

CC BY-SA 4.0
2022-01-17T22:31:41.947Z
For a rapidly-rotating star, your azimuth equation might be off by a bit, but this formula is great for the Sun because it's almost perfectly spherical: https://www.theguardian.com/science/2012/aug/16/sun-perfect-sphere-nature
giardia
CC BY-SA 4.0
2022-01-14T10:35:31.387Z
2022-01-14T13:36:49.730Z
2022-01-14T13:36:49.730Z
user45000
<p>I am reading about <a href="https://en.wikipedia.org/wiki/Limb_darkening" rel="nofollow noreferrer">limb darkening</a> and I am very confused how to find the value of <span class="math-container">$\mu$</span>.</p>
<p>It states it can be found by <span class="math-container">$\mu = \cos(\gamma)$</span>, but I am lost how <span class="math-container">$\gamma$</span> is found, or what it represents.</p>
<p>If a planet is, say in mid-transit, would <span class="math-container">$\gamma$</span> be zero degrees?</p>
<p>I can't seem to visualise the trigonometry to understand it. Hope someone can explain.</p>

CC BY-SA 4.0
2020-12-31T04:54:27.970Z
Can you add a link to the material you're reading?
Mike G
CC BY-SA 4.0
2020-12-31T07:42:52.483Z
The following questions and their answers and comments may be helpful. If you figure out the answer to your question on your own please feel free to post it as an answer! [Need help simulating solar limb darkening](https://space.stackexchange.com/q/16622/12102) and [Is the Sun slightly blue in the center? - Wavelength-dependent limb darkening of the Sun](https://astronomy.stackexchange.com/q/14952/7982) and [What is the intensity distribution of visible light over the solar disk?](https://astronomy.stackexchange.com/q/33342/7982)
uhoh
CC BY-SA 4.0
2020-12-31T07:43:16.553Z
and [Linear limb darkening coefficient, u](https://astronomy.stackexchange.com/q/12331/7982) btw are you reading [Wavelength dependency of the Solar limb darkening](http://www.physics.hmc.edu/faculty/esin/a101/limbdarkening.pdf)?
uhoh
CC BY-SA 4.0
2020-12-31T14:38:58.257Z
I am guessing that model 3 presented in Jeremy Tatums's notes *Simple Models of the Atmosphere to Explain Limb Darkening* at https://phys.libretexts.org/Bookshelves/Astronomy__Cosmology/Book%3A_Stellar_Atmospheres_%28Tatum%29/06%3A_Limb_Darkening/6.02%3A_Simple_Models_of_the_Atmosphere_to_Explain_Limb_Darkening is defining what the $\cos \gamma$ is about.
B--rian
<star><the-sun><planet><planetary-transits>
How to calculate the limb darkening $\mu$ value
WDUK
# How to calculate the limb darkening $\mu$ value

(Asked by: username_0 on 2020-12-31)

I am reading about [limb darkening](https://en.wikipedia.org/wiki/Limb_darkening) and I am very confused how to find the value of $\mu$.

It states it can be found by $\mu = \cos(\gamma)$, but I am lost how $\gamma$ is found, or what it represents.

If a planet is, say in mid-transit, would $\gamma$ be zero degrees?

I can't seem to visualise the trigonometry to understand it. Hope someone can explain.

**Tags:** star, the-sun, planet, planetary-transits

> **Comments:**
> **username_1**: Can you add a link to the material you're reading?  
> **username_2**: The following questions and their answers and comments may be helpful. If you figure out the answer to your question on your own please feel free to post it as an answer! [Need help simulating solar limb darkening](https://space.stackexchange.com/q/16622/12102) and [Is the Sun slightly blue in the center? - Wavelength-dependent limb darkening of the Sun](https://astronomy.stackexchange.com/q/14952/7982) and [What is the intensity distribution of visible light over the solar disk?](https://astronomy.stackexchange.com/q/33342/7982)  
> **username_2**: and [Linear limb darkening coefficient, u](https://astronomy.stackexchange.com/q/12331/7982) btw are you reading [Wavelength dependency of the Solar limb darkening](http://www.physics.hmc.edu/faculty/esin/a101/limbdarkening.pdf)?  
> **username_3**: I am guessing that model 3 presented in Jeremy Tatums's notes *Simple Models of the Atmosphere to Explain Limb Darkening* at https://phys.libretexts.org/Bookshelves/Astronomy__Cosmology/Book%3A_Stellar_Atmospheres_%28Tatum%29/06%3A_Limb_Darkening/6.02%3A_Simple_Models_of_the_Atmosphere_to_Explain_Limb_Darkening is defining what the $\cos \gamma$ is about.  

---

## 2 answers

### Answer by username_4 on 2022-01-17. Score: 1

* $\gamma$ = emission angle, the direction from a point on the surface of the limb-darkened body (presumably a star) toward the observer, with respect to the local vertical direction (0° at center of disk, 90° at edge of disk)
* $\mu$ = cos($\gamma$) = the emission angle cosine

If I infer that you're asking about the limb darkening of a star that is being transited by a planet, then neither the star nor the planet is resolved. But as the planet moves across the star, it will block the same angular area at different parts of the transit, but the brightness of the spot it's blocking will vary because of limb darkening. So this causes the [transit lightcurve to have a non-flat bottom](https://www.hao.ucar.edu/research/stare/hd209458.html).

A transit seen perfectly edge-on will have the planet blocking $\gamma$ = 0° at the midpoint of the transit, like you say in the original question. But usually transits are seen at some finite inclination, and so the planet blocks $\gamma$ = 90° at the start of the transit, but only reaches $\gamma\_{\rm{max}}$ at the midpoint, where 0° < $\gamma\_{\rm{max}}$ < 90°. The value of $\gamma\_{\rm{max}}$ will depend on the orbital radius, stellar diameter, and viewing inclination angle.


---

### Answer by username_5 on 2022-01-14. Score: 1

To calculate $\mu$ of a point on the Sun:
$\text{azimuth of point} = \arcsin(\frac{d}{r}),
\mu = \cos(\text{azimuth of point})$

where d=distance of point from centre of disc, and r is the radius of the disk.

You can try visualising it from the side of the disc. Here's a nice figure (though I would exchange the r and a notation for more clarity, as r usually denotes radius): <https://phys.libretexts.org/Bookshelves/Astronomy__Cosmology/Book%3A_Stellar_Atmospheres_(Tatum)/06%3A_Limb_Darkening/6.01%3A_Introduction._The_Empirical_Limb-darkening>

Mu will be 1 if the point is in disc centre (i.e., azimuth is 0 deg.), and it will be 0, if it is on the limb (i.e., azimuth is 90 deg.). So what it tells you basically is how far the point is from disc centre. That's why it is used to describe limb darkening.

I am not from the community that looks at planet transits, but I guess one could describe that also with mu, and then if it is right in the centre of the observed disc, azimuth would be indeed 0 degrees. However, as I understand, it should be referring to something on the surface, rather than something passing in front of the disc.

> **Comments:**
> **username_4**: For a rapidly-rotating star, your azimuth equation might be off by a bit, but this formula is great for the Sun because it's almost perfectly spherical: https://www.theguardian.com/science/2012/aug/16/sun-perfect-sphere-nature  

---

