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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?

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It's explained in its Wikipedia article: en.wikipedia.org/wiki/Chandrasekhar_limit –  Kvothe Feb 16 at 8:40

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up vote 3 down vote accepted

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 $M_{ch}=1.44M_\odot$.

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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. –  Warrick Mar 21 at 18:58

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