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?


2 Answers 2


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.

  • 1
    $\begingroup$ Thank you for the handwaving argument. The lecture notes will need some more time to digest. Your clear and to-the-point answers to innumerable questions in SE are an extra motivation for me to start as a regular astronomy student in university next year, at 67 years of age. $\endgroup$
    – gamma1954
    Commented Feb 1, 2021 at 18:02

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 $\alpha \geq 1$.

  • $\begingroup$ @gamma1954 If you, or any reader of this post, are interested into the full derivation - starting with the two fundamental equations -, I happy to add them upon your request to my answer. I just skipped it for the sake of succinctness. $\endgroup$ Commented Aug 27, 2020 at 16:38
  • $\begingroup$ Please read Rob Jeffries' answer to the previous question physics.stackexchange.com/questions/544435/… He mentions the counter-intuitive relation. I haven't yet seen a valid explanation in physical terms. $\endgroup$
    – gamma1954
    Commented Aug 28, 2020 at 17:29
  • $\begingroup$ Oh that odd and honestly, I haven't come across the claim that central density would indeed behave reciprocally to the total mass of the star. The only reason I can come up off the cuff would be that with the collapse of a larger molecular cloud, i.e. birthing a more massive star the potential energy converted into kinetic/thermal energy surpasses the thermal energy obtained by a smaller star by far. As a result, the central temperature would be substantially higher, and consequently a lower pressure/density in the core is needed to jump-start nuclear fusion. $\endgroup$ Commented Aug 29, 2020 at 12:53
  • $\begingroup$ Since you seek a valid explanation in physical terms, are you looking for a intuitive explanation for this behaviour or rather an argument based on physical laws and equations respectively? $\endgroup$ Commented Aug 29, 2020 at 12:56
  • $\begingroup$ I'm looking for an explanation using physical laws and relations, possibly with equations and/or diagrams. $\endgroup$
    – gamma1954
    Commented Aug 30, 2020 at 16:32

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