2
$\begingroup$

I guess this is more a question about stellar models than anything else. I was wondering what is predicted to be the hottest possible stars that would still be hydrogen-burning.

What complicates this more is that the most massive stars will probably only sit close to the main sequence (as very early O-type) for several thousand years before gaining a WNh spectrum despite still being relatively early in the lifetime. So I'll be somewhat more precise and ask what is the hottest possible star predicted at ZAMS that won't just blow itself apart by radiation pressure.

Also tangential to the topic but have there been any stars found with an O1 spectrum, and is this even a spectral type that has models made for it?

$\endgroup$
3
  • $\begingroup$ astronomy.stackexchange.com/questions/11594/… is basically a duplicate unless you don't know how hot an O3 star is. (About 50,000 K) $\endgroup$
    – ProfRob
    Jun 3, 2019 at 15:35
  • $\begingroup$ So ~50,000 K is about as hot as a ZAMS star can get? Is there a theoretical reason for this or have there just not been any stars observed with a hotter surface temp? $\endgroup$
    – ShroomZed
    Jun 3, 2019 at 16:20
  • 1
    $\begingroup$ The theoretical reason for pop I and II stars is likely that T_eff starts to drop as L goes higher, due to the presence of an optically thick stellar wind. $\endgroup$
    – Ken G
    Jun 6, 2019 at 16:39

2 Answers 2

8
$\begingroup$

An answer to your question is contained within What is the largest hydrogen-burning star? The hottest observed main sequence stars are of type O3V, with photospheric temperatures of about 50,000 K.

However, it is indeed possible that hotter main sequence stars may exist in the present-day universe, but have simply evolved into Wolf-Rayet stars (and lost a lot of mass). Indeed, Crowther et al. (2010) claim to see evidence for such objects in the cluster R136 in the Large Magellanic Cloud.

How hot could such objects be? A theoretical study by Bromm et al. (2001) suggests that very massive, metal-rich present-day stars might reach 65,000 K at a few hundred solar masses (see their Fig.1).

The hottest zero age main sequence stars ever would probably be very massive ($\sim 1000M_{\odot}$) ultra-metal poor, or even metal-free population III stars. These actually begin life as He burning stars until they produce enough carbon to commence the CNO H-burning cycle. Equation 6 and Fig.1 in Bromm et al. (2001), who provide a theoretical study of such objects, suggests effective temperatures of about $1.1\times 10^{5}$ K for such stars, with very little detailed dependence on mass and metallicity.

I guess such a star might be classified as O1, but according to Bromm et al. they are basically blackbodies once the mass exceeds about 300 solar masses.

$\endgroup$
5
  • 1
    $\begingroup$ I've never heard of stars starting at a He-burning phase so that's extremely interesting. Thank you for taking the time to state this. $\endgroup$
    – ShroomZed
    Jun 3, 2019 at 16:45
  • $\begingroup$ @ShroomZed It makes sense, though. The first stage of the p-p chain is really slow because the probability of diprotium converting to deuterium is really small. Most of the time, it quickly falls apart into a pair of protons. $\endgroup$
    – PM 2Ring
    Jun 3, 2019 at 20:36
  • $\begingroup$ Yeah, I'm aware of that. I've heard that for every 10^26 times two protons interact in the core, only one successful reaction occurs in the sun. For a massive metal free star, CNO cycle is the only way to go. $\endgroup$
    – ShroomZed
    Jun 3, 2019 at 21:32
  • 3
    $\begingroup$ The reason is that the pp-chain temperature dependence is sufficiently low that the temperature reaches He-burning temperatures before the collapsing star can be stabilised. Once the C abundance reaches a threshold, the CNO cycle exceeds the He burning rate. @ShroomZed $\endgroup$
    – ProfRob
    Jun 3, 2019 at 21:58
  • $\begingroup$ On a different note, this also means that the most massive stars in R136 must have started out with surface temps above 60,000 K. $\endgroup$
    – ShroomZed
    Jun 6, 2019 at 16:22
2
$\begingroup$

Another key point to bear in mind is that if one is asking for the "hottest" star, then presumably it is "effective temperature" that is referenced. Effective temperature is related to the luminosity L and the radius R from which the light emerges (not the static surface of the star, R can be well out in a dense stellar wind) by the Stefan-Boltzmann law, which says T_eff ~ L^(1/4) / R^(1/2). Normally for main-sequence stars, whose R rises much more slowly than L, that means the T_eff is higher at higher L. But really high L stars have dense radiatively-driven winds, and eventually the wind could get so dense that the mass-loss rate from the star might be proportional to L. If so, the radius R from which the light emerges (where the optical depth is near unity) is also proportional to L, so then T_eff ~ L^(1/4) / L^(1/2) ~ L^(-1/4).

This says we should expect T_eff to reach a peak and then drop if L is further increased. The peak T_eff should be when the winds are getting so dense that the mass-loss rate becomes proportional to L (when a significant amount of L is being used to lift the mass), and that is more or less also where the WNh spectrum replaces the O spectrum (something like what is trying to be the O1 spectrum). So the kinds of T_eff already referred to, above 50,000 K or so, is going to be the highest T_eff possible for a core hydrogen burning star. After that, even higher L will simply mean that you see a much larger star with a lower T_eff.

Incidentally, this probably points to a flaw in the Bromm et al. paper cited above. That paper is really about Pop III stars, so zero metallicity and not much in the way of stellar winds also. So when they talk about T_eff, they appear to use an expression that involves L and the static radius, but that's not the T_eff that matters for observing stars, or for their UV flux. The latter should use the R from which the light emerges, which can be in the wind. The distinction is not important for the pop III conclusions of their paper, but it does matter for the Pop I results shown in Figure 1. That is probably not the correct T_eff, so I doubt T_eff really can get up to 65,000 K for Pop I stars.

$\endgroup$
2
  • $\begingroup$ Massive Pop III MS stars do not have zero metallicity. They have already undergone a brief He burning phase. $\endgroup$
    – ProfRob
    Jun 6, 2019 at 17:50
  • $\begingroup$ Not in the core, that's true. I'm referring to the wind, because that's what matters to T_eff. We can assume the internal structure is dominated by core CNO cycle H-fusion, but what the star looks like will also be controlled by its wind. The Pop III stars will have zero metallicity winds, which makes their winds very weak. But Pop I will have strong winds, and this does not appear to be included in Figure 1. $\endgroup$
    – Ken G
    Jun 6, 2019 at 18:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .