Why does a star beyond a certain mass limit (Chandrasekhar limit) only become a black hole?

A star is first made of hydrogen, it undergoes nuclear fussion reaction combining into helium and releasing a large amount of energy. This process continues till star is made up of iron core as iron has largest value of binding energy per nucleon, after this if mass of the star is above the value of Chandrasekhar limit it becomes a black star, what is the reason for this and why is certain mass limit required?


2 Answers 2


You are a little confused in your stellar evolution model. After the ignition of hydrogen fusion in the core of a star, it will next progress to helium fusion, then to carbon/oxygen fusion via the triple-alpha process (I've skipped a lot of steps and details there, if you want the details you can look at either Hansen & Kawaler's Stellar Interiors text or Dina Prialnik's Introduction to Stellar Structure text). What happens next is mass-dependent (using $M_\odot\simeq2\cdot10^{33}$ g and the mass of the star as $M_\star$):

  • $M_\star\gtrsim 8M_\odot$
    • able to continue fusion in the core
    • will later blow up in core-collapse supernova events, producing either a neutron star or a black hole (mass-dependent) after forming iron in the core
  • $M_\star\in(\sim0.5,\,\sim8)M_\odot$
    • unable to continue fusion in the core due to insufficient temperatures
    • will proceed into the planetary nebula phase (which has nothing to do with forming planets, but it's discoverer, William Herschel, thought that it was a planetary system forming)
    • these stars form the white dwarfs that the Chandrasekhar limit applies to
  • $M_\star\lesssim0.5M_\odot$
    • unable to produce helium in the core (insufficient temperatures)
    • expected to continue burning hydrogen for $t_{burn}>t_{age\,of\,universe}$

Thus, not every star produces iron in the core; this only applies to stars with mass $\gtrsim8M_\odot$.

The Chandrasekhar limit arises from comparing the gravitational forces to an $n=3$ polytrope (see this nice tool from Dr Bradley Meyer at Clemson University on polytropes)--polytropes basically mean $P=k\rho^{\gamma}$ where $P$ is the pressure, $k$ some constant, $\rho$ the mass density and $\gamma$ the adiabatic index.

That is, in order to find the limit, you need to use the hydrostatic pressure, $$ 4\pi r^3P=\frac32\frac{GM^2}{r}\tag{1} $$ and insert the pressure of the polytrope of index $n=3$ (requires numerically solving the Lane-Emden equation) and then solving (1) for the mass, $M$. If you've done it correctly, you'll find $M_{ch}=1.44M_\odot$.

  • 2
    $\begingroup$ I would only add that in this case, the use of an $n=3$ polytrope is well justified, since it's what you get for electron degeneracy in the limit that all the velocities tend to the speed of light. $\endgroup$
    – Warrick
    Mar 21, 2014 at 18:58
  • $\begingroup$ So that means all the iron on Earth came from supernova event(s) from a star with mass $M_\star\gtrsim 8M_\odot$, right? $\endgroup$
    – user541686
    Aug 14, 2014 at 10:22
  • $\begingroup$ @Mehrdad: that is pretty much true for all elements that are heavier than iron. $\endgroup$
    – Kyle Kanos
    Aug 14, 2014 at 10:49
  • $\begingroup$ 1.44 solar masses is not the Chandrasekhar limit for an iron core. $\endgroup$
    – ProfRob
    Oct 13, 2014 at 21:03
  • 1
    $\begingroup$ @KyleKanos Not so. The Chandrasekhar limit can be defined in terms of an $n=3$ ideal polytrope, but the composition of the gas, particularly the number of mass units per electron, $\mu_e$, is crucial ($k$ in your answer is not a constant). The "traditional", Newtonian value is $M_{Ch}=1.44(\mu_e/2)^{-2}$, where $\mu_e=2.15$ for 56Fe. Chandrasekhar used $\mu_e=2.5$ in his original paper and got $M_{Ch}=0.91$ ! ( adsabs.harvard.edu/abs/1931ApJ....74...81C ). There is no debate that GR ought to be included and this lowers $M_{Ch}$ further. No deviations from polytropes are required. $\endgroup$
    – ProfRob
    Oct 26, 2014 at 20:41

The Chandrasekhar mass is not the dividing line between those stellar remnants that will become black holes and those that will become something else.

A compact, cold white dwarf (i.e. one supported by electron degeneracy pressure) may become unstable and collapse at close to the value of $M_{Ch}=1.44(\mu_e/2)^{-2}M_{\odot}$, where $\mu_{e}$ is the number of mass units per free electron ($\mu_e=2$ for Carbon or Oxygen) and derived using simple Newtonian mechanics. [In fact the Chandrasekhar mass is likely lower because of (i) electrostatic Coulomb corrections to the equation of state; (ii) inverse beta decay inducing instability and/or (iii) General Relativistic instability at finite density]. Anyhow, it's probably between 1.3 and 1.4 solar masses for a carbon/oxygen WD. If a white dwarf gained more mass than this it would either explode as a type Ia supernova or collapse to form a stable neutron star (e.g. Fryer et al. 1999); and certainly would not form a black hole.

The scenario described in the question is of a star forming an iron core. In this case $\mu_e =56/26$ and $M_{Ch}$ calculated from ideal electron degeneracy pressure is more like 1.24$M_{\odot}$ and reduced even further to 1.06$M_{\odot}$ by inverse beta decay instability. (e.g. Boshkayev et al. 2018).

If the core exceeds this value it will collapse, but that does not mean a black hole will form. The most likely outcome, at least for progenitor masses $<20-30M_{\odot}$ may be the formation of a neutron star supported by the strong nuclear repulsion between closely-packed neutrons. The dividing line between those objects that become black holes and those that become neutron stars is highly uncertain and may depend strongly on other factors like rotation.

  • $\begingroup$ Note also that rotating white dwarfs can also reach north of 1.5 solar masses (I've seen as high as 5 in a conference proceeding that I cannot locate at the moment). $\endgroup$
    – Kyle Kanos
    Oct 14, 2014 at 0:29
  • $\begingroup$ @KyleKanos Please supply the reference if you can find it. Here is what I think is the consensus view adsabs.harvard.edu/abs/2013ApJ...762..117B The Chandrasekhar mass is never more than 6% larger than for non-rotating WDs. Note that GR does not permit compact stars to exceed 3.5 solar masses (with any equation of state or rotation). $\endgroup$
    – ProfRob
    Oct 14, 2014 at 8:48
  • $\begingroup$ Another factor that determines whether a collapsing object forms a NS or a BH is the metallicity of the progenitor. i.e. see figure 1 of arxiv.org/pdf/astro-ph/0212469.pdf $\endgroup$ Feb 5, 2019 at 21:08

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