I wonder whether e.g. stars more massive than the Sun (but below the mass where they'd go supernova) would become larger or smaller white dwarfs than a white dwarf Sun. Since Procyon B is larger than the more massive Sirius B, I suppose the more massive the star the smaller its remnant white dwarf would become. Am I right and if so, why is it like this?


More massive stars have a more massive core and produce more massive white dwarfs. The relationship between the initial mass of the main sequence star and the final mass of the white dwarf is monotonic, but not linear. The Sun is expected to produce a white dwarf with a mass of around $0.5 M_\odot$, whilst a $8M_\odot$ star is expected to produce a carbon/oxygen white dwarf of mass $\sim 1.1 M_\odot$. It is possible that some stars with $8-9M_\odot$ could result in a slightly more massive oxygen/neon/magnesium white dwarf.

The difference between the initial and final masses is accounted for by mass loss in the red giant and asymptotic giant branch phases. Note that white dwarfs close to the Chandrasekhar limit are probably not produced by the normal evolution of a single star and that stars with initial mass $>9M_\odot$ probably end their lives with a supernova of some kind, that either destroys the star or leaves behind a neutron star or black hole.

White dwarf stars are supported by electron degeneracy pressure. This form of pressure, does not depend on temperature (to first order), but does depend on density. To work out how big (in terms of radius) a white dwarf of a given mass is, one uses the density-dependent pressure to solve the hydrostatic equilibrium equation and one finds that for lower densities (which means $<10^{10}$ kg/m$^3$ in this context!), the pressure $P$ is proportional to density $\rho^{5/3}$, and this results in a radius $R$ that is proportional to mass $M^{-1/3}$. That is, a more massive star has a smaller radius. The increased weight of a more massive star requires a bigger pressure gradient to support it and this is provided by a bigger central pressure and smaller radius.

The mathematics of this is a very standard piece of bookwork involving "polytropes" (where the pressure is proportional to density to some power) and I won't produce it. A simple argument is the following. The hydrostatic equilibrium equation is $$ \frac{dP}{dr} = - \rho g,$$ where $g$ is the gravity due to the mass within radius $r$.

If we work in proportionality, then $P \propto \rho^{5/3} \propto M^{5/3} R^{-5}$. Thus $dP/dr \propto M^{5/3} R^{-6}$.

On the right hand side, $\rho \propto MR^{-3}$ and $g \propto MR^{-2}$.

Thus the hydrostatic equilibrium equation tells us $$ M^{5/3}R^{-6}\propto M^2 R^{-5}$$ and so $$ R \propto M^{-1/3}.$$

As a more massive white dwarf gets smaller, the density goes up as $\rho \propto M^2$ and the degenerate electrons become relativistic. This drives the pressure equation towards $P \propto \rho^{4/3}$. This leads the radius to decrease more steeply as a function of mass and it asymptotes to zero radius and infinite density at the Chandrasekhar mass.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.