We know that Vega is the star that serves as the zero point for the UBV color scale, and has an apparent magnitude of nearly zero (+0.02). But its absolute magnitude is +0.58, making it rather far from $M=0$. So what spectral type would have an absolute magnitude of $0 \pm0.02$, and are there any stars that satisfy this?
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4$\begingroup$ How precisely do you define an absolute magnitude of exactly 0? The more exactly such a star is required to have an absolute magnitude of exactly 0, the fewer the stars will be that have such a magnitude. The more exactly such a star is required to have an absoluted magnitude of exactly 0, the fewer the stars will be which have their absolute magnitudes measured and calaculated that precisesly, until eventually an increasing demand for a precise magnitude will lead to no stars having such precise magnitude known. $\endgroup$– M. A. GoldingNov 4, 2020 at 16:49
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4 Answers
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There isn't a one-to-one relationship between spectral type and absolute magnitude.
Instead, there is a mean relationship with a fair bit of scatter around it. The reason is that the luminosity of a star of a given effective temperature depends on its composition/metallicity and how far along in its main sequence lifetime it is.
Basically, late B-type main sequence stars (say B7/B8V) have an absolute magnitude of about zero. Alternatively there are low mass stars ascending the hydrogen shell burning giant branch (types of about K2-K5 III) that would have an absolute magnitude of zero.
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A star with magnitude 0 would be 85 times brighter than the sun (since Magnitude=-2.5 log(Luminiosity))
Referring to the H-R diagram on Wikipedia shows that there is quite a range of spectral types possible with this luminosity: from B type main sequence stars, and A type sub-giants, such as 4 Sco
There are also G and K and M type Giant stars with this luminosity. such as ρ UMa
Identifying stars that are exactly this bright is made more difficult by the difficulty in measureing the distance to stars.
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According to Eric Mamajek's table of main sequence stars, the absolute V-magnitude of zero corresponds to late B-type stars. The values that bracket zero are B8V with $M_{\rm v} \approx -0.2$ and B9V with $M_{\rm v} \approx 0.7$. There is a fair amount of scatter around the main sequence so it is likely that some stars a few spectral subtypes away may have absolute magnitudes near zero. Note that the table there only covers the main sequence, it does not cover other luminosity classes that may also fall into this range, e.g. subgiants and giants.
To find stars with absolute magnitudes close to zero, you can for example query the XHIP catalogue via VizieR and enter a range, e.g. -0.02 .. 0.02 in the "VMag" field (not the "Vmag" field, as the one with lower-case m is apparent not absolute magnitude), and get a list of Hipparcos stars with absolute magnitudes calculated as being near zero. The first ones that come up are HIP 63 (A0IIspSiSrHg), HIP 422 (K0III) and HIP 602 (A3V (shell)). You need to watch out though as this list may contain unresolved binary stars.
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$\begingroup$ Ah shit, didn't realise VMag and Vmag was different. No wonder everything looked wrong. Thanks for pointing that out, that bit of info right there spared my sanity. $\endgroup$ Aug 23, 2021 at 2:31
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The absolute magnitude quantifies the luminosity of an object at a standard distance of $10\ \mathrm{pc}$ from Earth. For example, in the case you mentioned, Vega becomes dimmer than at its actual distance (about $7\ \mathrm{pc}$).
To answer your question, I don't think there is an actual star with exactly 0 absolute magnitude. If there is then, following the formula of the absolute magnitude, where $d_{pc}$ is the actual distance of the star from Earth in parsecs:
$$M=m-5\log_{10}\left(\frac{d_\mathrm{pc}}{10\ \mathrm{pc}}\right)$$
Then such star should satisfy the following relation between apparent magnitude and distance to have absolute magnitude zero: $$\frac m5+1=\log_{10}(d_\mathrm{pc})$$
E.g: if you have a star at the Vega's distance $d_\mathrm{pc}=7.63\ \mathrm{pc}$ with apparent magnitude $m\simeq-0.58$ (more or less as Saturn maximum apparent magnitude) then you will have $M=0$
