Relation between central density and stellar mass (2) Comparing stars with different masses, the central density is lower in a heavy star than in a low mass star (assuming that each star has the same composition and has just reached the stage in which it is mostly powered by hydrogen fusion). The lower central density in a heavier star seems counter-intuitive to me.
How can this relation be explained in terms of physical mechanisms?
 A: You can do something rough using homology relations which assume homologous structures for stars governed by a similar polytropic relation between pressure and density. So this will work for stars like the Sun and more massive where we can assume energy is transported radiatively and the structure of the star is governed by a $n=3$ polytrope.
The derivations are still quite lengthy, I'm not going to reproduce them; you can find them in this set of lecture notes for example.
The basic result is that the central density scales like
$$\rho \propto M^{2(3-\nu)/(3+\nu)}\ ,$$
where $\nu$ is the power law index governing the sensitivity of the nuclear burning rate to temperature: $\nu \sim 4$ for pp burning and $\nu \sim 20$ for the CNO cycle.
Thus $\rho \propto M^{-2/7}$ for pp burning stars with mass less than about $1.5M_\odot$, but density falls more steeply as $\rho \propto M^{-34/23}$ in more massive CNO burning stars.
Short of going through the stellar structure equations (which the link above does), a handwaving physical argument would go as follows.
The virial theorem easily tells us that the interior temperature of a star $ T \propto M/R$.
If the energy generation rate is strongly dependent on temperature and steeper than the mass-luminosity relationship, which it certainly is for the CNO cycle, then the central temperature hardly changes with mass and so $R \propto M$.
From this we see that density scales as $M^{-2}$.
In practice, the central temperature does rise slightly with mass, so $R$ does not quite rise linearly with mass and so the density decrease is not quite as steep. For the pp cycle, $\nu \sim 4$ is only slightly steeper than the luminosity-mass relation, so central temperature rises significantly with mass and hence the central density only falls slowly with mass.
A: Unfortunately, I am not aware of such an indeed counter-intuitive relation. I also conducted a brief (thus definitely not comprehensive) search on the web and I could only find sources that back up the intuitive, opposite relation (e.g., https://home.strw.leidenuniv.nl/~nielsen/SSEs16/Lectures-2016/Lecture-9-3_2016.pdf slide no. 7).
In fact, by applying two out of the four stellar structure equations, namely the mass continuity equation and the hydrostatic equilibrium equation, one can show that - at least in theory (no one has every measured the pressure of low- or high-mass stars in situ) - the central pressure $P_c$ is directly proportional to the mass of the star $M_\star$,
\begin{align}
    P_c \equiv P_c(M_\star,R) = \frac{5}{4 \pi} \, \frac{G \, M_\star^2}{R^4} \quad.
\end{align}
Moreover, on can also show that the central pressure $P_c$ is directly proportional to the central density $\rho_c$ (which is immediately intuitive in my opinion). So, with using the basic equations of stellar structure it can be deduced that,
\begin{align}
    \boxed{\; \rho_c \propto M_\star^{\alpha} \;} \quad,
\end{align}
with $\alpha \geq 1$.
