# Why do massive rapid-rotating stars move bluewards off the zero-age main sequence (ZAMS)?

I'm reading a well-cited paper by Szecsi et al. 2015 titled "Low-metallicity massive single stars with rotation". The paper can be found here. I was surprised that figure 5 depicted an HRD in which rapid rotators of 500 km/s evolved bluewards on the HRD. I didn't know any star could move anything but rewards when leaving the ZAMS. Other rotators below that 500 km/s move in a more typical fashion. Can anyone explain what is going on here?

• You need to explain what the lines on the graph represent. Feb 23 at 19:18
• The paper attributes this to chemically homogeneous evolution caused by rotational mixing. Did you look at the Maeder (1987) paper it references? Feb 23 at 19:28
• @ ProfRob: I updated the plot, thanks. I see that the mixing causes it to be more homogenous but I'm not sure how that changes the temperature. It's not obvious to me. Feb 23 at 21:45

The blueward evolution is a simple consequence of the homology relations for main sequence evolution.

If the star is rotating very fast, then this induces strong mixing of the interior, such that the star can be treated as chemically homogeneous. As main sequence hydrogen burning proceeds, the entire star is gradually turned into helium, with a consequent increase in the mean molecular weight $$\mu$$.

Therefore, the evolution can be thought of as a sequence of models with similar mass, but increasing $$\mu$$.

The luminosity evolution is $$L \propto \mu^{7.5} M^{5.5} R^{-0.5}\ .$$ The strong $$\mu$$ dependence comes about through the strong dependence of opacity on $$\mu$$ for stars in radiative equilibrium.

On the other hand, the radius is only weakly dependent on $$\mu$$. That is because the central temperature depends on $$\mu M/R$$ (from hydrostatic equilibrium), but the strong temperature dependence of the CNO cycle, means the central temperature hardly increases as burning continues. As a result $$R \propto \mu^{2/3} M^{4/5}\ .$$

If we assume a blackbody, then $$T_{\rm eff}\propto L^{1/4}R^{-1/2}$$ and thus $$T_{\rm eff}\propto \mu^{1.54} M^{0.98} R^{-0.12}\ .$$

If we start off with 10% He by number, then $$\mu=0.56$$. But if we end with 80% He, then $$\mu=1.2$$.

On the face of it, this would appear to more than double the surface temperature during the main sequence evolution. The effect will be mitigated by mass loss and the disproportionate effect on opacity contributed by whatever metals are present (these are not zero metallicity stars).