2
$\begingroup$

I am reading about 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.

$\endgroup$
4
0
$\begingroup$

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.

$\endgroup$
1
0
$\begingroup$
  • $\gamma$ = 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.

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.

$\endgroup$

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.