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One hears that hyper-giant stars like Eta-carinae represent the upper bound on stellar mass at about 130 solar masses. At the lower end there are red dwarfs with as little as 0.075 solar mass.

You could say, then, that Eta-carinae is somewhere up in the 99.99th centile of our sample of known stellar masses, while a red dwarf with 0.075 solar mass is somewhere around the 0.01st centile.

I am curious as to the distribution in between. What stellar mass centile does the sun lie in? What is the median stellar mass? Is there a parametric model which best fits the distribution? If astronomers have not yet come up with a stellar mass distribution model, why not? Is it considered uninteresting?

David Z
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ben
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1 Answers1

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You are looking for the initial mass function (IMF). This tells the probability of finding a star of mass $m$ (in solar mass units). The prototypical IMF is the Salpeter IMF, $$ \phi(m)\sim m^{-2.35} $$ This gives a decent and quick approximation, though a multi-power-law fit seems to be better; thus we have the Kroupa 2001 model: $$ \phi(m)\sim \begin{cases}m^{-0.3} & \qquad \quad m<0.08 \\ m^{-1.3} & 0.08\leq m<0.5 \\ m^{-2.3} & \qquad \quad m\geq0.5\end{cases} $$ You should be able to calculate the centiles from this, $$ N(m)=\int\phi(m)\,dm $$ (since they're fairly simple integrals, why not compare both models?). The upper and lower bounds are often taken to be 0.01 and 150 respectively, though these values aren't really set in stone.

The mean stellar mass comes from $$ \langle m\rangle=\int m\phi(m)\,dm, $$ which is, again, an easy integral to perform. The median stellar mass comes from finding $m_{med}$ such that $$ \int_{m_{min}}^{m_{med}}\phi(m)\,dm=\frac12 $$

For a bit more information on the IMF, see this answer (it touches on the lower-limit of the IMF and some open questions on the IMF).

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Kyle Kanos
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