<p>It takes a spot &quot;bump&quot; half the rotation period to fully traverse from the blue side of the line profile to the red.</p>
<p>The bump tracks the position of the spot in velocity. The blue side of a line profile is produced by the limb of the star approaching the Earth, where the resolved component of the velocity towards Earth is largest. The red edge of the profile is the opposite limb that is receding from Earth (compared with the blue side). When the spot produces a bump in the blue side of the profile, that means the spot is near the approaching limb. When the bump is on the red edge of the profile, the spot is on the receding limb. Since we can see half the surface of the star at a time, it takes half a rotation period for the spot to cross from one limb to the other.</p>
<p>You can see that the bump has maybe moved across 20% of the profile in one night, so the rotation period could be estimated as <span class="math-container">$\sim 10$</span> days.</p>
<p>The rotation period of Sigma Gem is 19.6 days. So my &quot;by eye&quot; method is not very accurate (or perhaps the spectra are more than 24 hours apart). I also assumed the spot tracks across the profile at a uniform rate, but that isn't quite right. The connection between spot velocity and phase is a sinusoidal function, so that the spot appears to move more quickly through the centre of the profile and slower near the edges. This could also account for my underestimate.</p>
<p>To get a more accurate estimate you would need to simulate the star plus spot spectrum to take account of limb darkening, projection effects and the spectral resolution of the spectrograph.</p>

CC BY-SA 4.0
2021-05-03T08:21:31.180Z
2024-09-04T07:49:18.320Z
2024-09-04T07:49:18.320Z
ProfRob
<p><a href="https://i.sstatic.net/FMhWU.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/FMhWU.png" alt="enter image description here" /></a></p>
<p>The figure is shown; the measurements were taken on two consecutive observing nights. The Ordinate is the flux normalized to continuum and the abscissa is the wavelength scale. You can see the &quot;bumps&quot; indicated by the arrows referring to some Starspot as the spot moves on the profile; assuming a single time-stable position-stable spot.</p>
<p>The &quot;bumps&quot; slightly shifts, as indicated by the arrows in the top line profile compared to the bottom profile, as the spot &quot;moves across&quot; the surface as the star as it rotates; i'm just not sure how to get an estimate of the rotational period from this? Given just the wavelength for the abscissa.</p>
<p>Figure reference : Gray, D. (2005). The Observation and Analysis of Stellar Photospheres (3rd ed.), page 498. Cambridge: Cambridge University Press. doi:10.1017/CBO9781316036570</p>

CC BY-SA 4.0
2021-05-04T11:09:16.377Z
Could be red / blu shit if the margins of the disk can be resolved. But no idea about the amount.
Alchimista
CC BY-SA 4.0
2021-05-06T07:58:08.307Z
As there are still no answers, I'll throw in my ideas. I'm not at all sure how the "large spots" cause the bumps in the spectrum, but somehow their shifting seems to be a Doppler effect. Then if the bumps are persistent over a time long enough, you can find the periodicity of the bump positions, directly giving the rotation period. And the amount of Doppler shift gives an estimate of the radial velocity, allowing you to calculate the star's diameter.
Ralf Kleberhoff
<rotation><spectroscopy><spectra><stellar-astrophysics><starspot>
How would one estimate the rotation period of a star from its spectrum?
Jay D
# How would one estimate the rotation period of a star from its spectrum?

(Asked by: username_0 on 2021-05-03)

[![enter image description here](https://i.sstatic.net/FMhWU.png)](https://i.sstatic.net/FMhWU.png)

The figure is shown; the measurements were taken on two consecutive observing nights. The Ordinate is the flux normalized to continuum and the abscissa is the wavelength scale. You can see the "bumps" indicated by the arrows referring to some Starspot as the spot moves on the profile; assuming a single time-stable position-stable spot.

The "bumps" slightly shifts, as indicated by the arrows in the top line profile compared to the bottom profile, as the spot "moves across" the surface as the star as it rotates; i'm just not sure how to get an estimate of the rotational period from this? Given just the wavelength for the abscissa.

Figure reference : Gray, D. (2005). The Observation and Analysis of Stellar Photospheres (3rd ed.), page 498. Cambridge: Cambridge University Press. doi:10.1017/CBO9781316036570

**Tags:** rotation, spectroscopy, spectra, stellar-astrophysics, starspot

> **Comments:**
> **username_1**: Could be red / blu shit if the margins of the disk can be resolved. But no idea about the amount.  
> **username_2**: As there are still no answers, I'll throw in my ideas. I'm not at all sure how the "large spots" cause the bumps in the spectrum, but somehow their shifting seems to be a Doppler effect. Then if the bumps are persistent over a time long enough, you can find the periodicity of the bump positions, directly giving the rotation period. And the amount of Doppler shift gives an estimate of the radial velocity, allowing you to calculate the star's diameter.  

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## 1 answers

### Answer by username_3 on 2021-05-03. Score: 2

It takes a spot "bump" half the rotation period to fully traverse from the blue side of the line profile to the red.

The bump tracks the position of the spot in velocity. The blue side of a line profile is produced by the limb of the star approaching the Earth, where the resolved component of the velocity towards Earth is largest. The red edge of the profile is the opposite limb that is receding from Earth (compared with the blue side). When the spot produces a bump in the blue side of the profile, that means the spot is near the approaching limb. When the bump is on the red edge of the profile, the spot is on the receding limb. Since we can see half the surface of the star at a time, it takes half a rotation period for the spot to cross from one limb to the other.

You can see that the bump has maybe moved across 20% of the profile in one night, so the rotation period could be estimated as $\sim 10$ days.

The rotation period of Sigma Gem is 19.6 days. So my "by eye" method is not very accurate (or perhaps the spectra are more than 24 hours apart). I also assumed the spot tracks across the profile at a uniform rate, but that isn't quite right. The connection between spot velocity and phase is a sinusoidal function, so that the spot appears to move more quickly through the centre of the profile and slower near the edges. This could also account for my underestimate.

To get a more accurate estimate you would need to simulate the star plus spot spectrum to take account of limb darkening, projection effects and the spectral resolution of the spectrograph.


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