The Earth is a very smooth sphere, and the Sun even more so, with only minor fluctuations. I am wondering: are larger stars even rounder? Intuitively, that seems self evident, but I am not so sure. For instance, the hydrostatic equilibrium causes larger stars to be much less dense than red dwarfs. So what is the most important factor for how round a star is, a higher mass, or less activity? The most prominent cause of irregularities is of course the rotation rate of the star, which is pretty much independent of size. Ignoring that, do larger stars have smaller a deviations from the ellipsoid relative to their size?


As it seems like the "other than rotation rate" criterion is not really meaningfull, I now terminate it.

  • $\begingroup$ What do you mean by rounder? $\endgroup$
    – ProfRob
    Commented Apr 29, 2016 at 19:04
  • $\begingroup$ Indeed, more round is smaller irregularities relative to size. $\endgroup$ Commented Apr 29, 2016 at 19:10
  • $\begingroup$ I have tried to edit the question now, to make it more clear I am asking about irregularities, not oblateness. $\endgroup$ Commented Apr 29, 2016 at 19:21
  • $\begingroup$ I thought larger stars tended to have higher angular velocity than smaller stars? $\endgroup$
    – called2voyage
    Commented Apr 29, 2016 at 19:25
  • 2
    $\begingroup$ I have no idea how you would define an irregularity for a gaseous spheroid. In any case, we have no images or measurements of this for stars other than the Sun. $\endgroup$
    – ProfRob
    Commented Apr 29, 2016 at 21:24

1 Answer 1


In terms of mean angular velocity, the distribution of rotation rates among main sequence stars is well known. Allen (1963) compiled data on mass, radius, and equatorial velocity, which was then expanded upon by McNally (1965), who focused on angular velocity and angular momentum. It became clear that angular velocity increases from low rates for spectral types of G and below before rising to a peak around type A stars and then slowly decreasing.

Equatorial velocity continues increasing to mid-B type stars, before slowly decreasing, but because of the increased radii of O and B type main sequence stars, the peak in angular velocity occurs before this. As part of Jean-Louis Tassoul's Stellar Rotation notes, many O type stars have rotational periods similar to that of the G-type stars like the Sun!

The distribution is not smooth and uniform (McNally noticed a strange discontinuity in angular momentum per unit mass right for A0 and A5 stars; see his Figure 2); Barnes (2003) observed two distinct populations in open clusters, consisting of slower rotators (the I sequence) and faster rotators (the C sequence). Stars may migrate from one sequence to another as they evolve. Interestingly enough, stars on the I sequence lose angular momentum $J$ faster than stars on the C sequence: $$\frac{\mathrm{d}J}{\mathrm{d}t}\propto-\omega^n,\quad\text{where}\begin{cases} n=3\text{ on the I sequence}\\ n=1\text{ on the C sequence}\\ \end{cases}$$ Here, of course, $\omega$ is angular velocity. These results obey Skumanich's law.

Oblateness can be determined from mass, radius, and angular velocity as $$f=\frac{5\omega^2R^3}{4GM}$$ Using this and McNally's data, some quick calculations get me the following table: |Spectral type|$f/f(O5)$| |--|-------| |O5 | 1 | |B0 | 1.28 | |B5 | 1.84 | |A0 | 1.67 | |A5 | 1.35 | |F0 | 0.482| |F5 | 0.0387| |G0 | 0.000314|

  • $\begingroup$ I realize my question perhaps is not answerable in the form it was asked. But this is awesome anyway. $\endgroup$ Commented Apr 30, 2016 at 6:39
  • $\begingroup$ What has this got to do with how smooth a star is? $\endgroup$
    – ProfRob
    Commented Apr 30, 2016 at 17:03
  • $\begingroup$ @Hohmannfan Does it still satisfy the criteria of the question? I didn't discuss irregularities at all. $\endgroup$
    – HDE 226868
    Commented Apr 30, 2016 at 21:52
  • $\begingroup$ @HDE226868 There was quite some discussion going on if my question was meaningful excluding angular velocity, so in the end, I decided to change that. (Your answer was what finally convinced me.) But if you have got some impossible-to-obtain stellar data, feel free to include it :) $\endgroup$ Commented Apr 30, 2016 at 22:08

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