2
$\begingroup$

enter image description here

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

$\endgroup$
2
  • $\begingroup$ Could be red / blu shit if the margins of the disk can be resolved. But no idea about the amount. $\endgroup$
    – Alchimista
    May 4 at 11:09
  • $\begingroup$ 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. $\endgroup$ May 6 at 7:58
1
$\begingroup$

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

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

Looking up the period of Sigma Gem - it is 19.6 days. So my method is not very accurate by eye, but you would need to simulate the star plus spot spectrum more accurately to take account of limb darkening and the spectral resolution of the spectrograph.

$\endgroup$
0

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.