21
$\begingroup$

After a star becomes a White dwarf, it resists gravitational collapse mainly due to the electron degeneracy pressure. If the mass of the white dwarf is greater than the Chandrasekhar limit, the degeneracy pressure cannot resist the collapse any longer and is doomed to become a neutron star or a black hole. Why can't the degeneracy pressure keep on self-adjusting itself to resist collapse forever?

$\endgroup$

2 Answers 2

37
$\begingroup$

The basic problem is that for a sufficiently massive star, the electrons become relativistic. The fine details of this calculation are rather complicated, but you can get a qualitative sense of the argument as follows:

For non-relativistic fermions at zero temperature, it is possible to show that the total energy of $N$ particles in a box of volume $V$ is proportional to $N^{5/3}/V^{2/3}$. This can be done via counting the density of states, and using the fact that the energy of a non-relativistic particle obeys $E \propto |\vec{p}|^{2}$. For a spherical volume of radius $R$, we have $R \propto V^{1/3}$, and the number of fermions present is proportional to the mass. This means that the total energy of the fermions is proportional to $M^{5/3}/R^2$. This energy is positive.

On the other hand, the gravitational energy of a solid sphere is negative and proportional to $M^2/R$. This means that the total energy is the sum of a negative $R^{-1}$ term and a positive $R^{-2}$ term, and such a function will have a minimum somewhere. This will be the equilibrium point. At smaller radii, the energy of the degeneracy grows faster than the binding energy decreases, pushing the radius back to larger values. At larger radii, the reverse occurs. This means that the star will be stable.

This argument doesn't hold up to arbitrarily large energies, though, because eventually the Fermi energy of the electrons exceeds the rest energy of the electron; in other words, the electrons become relativistic. This changes the relationship between energy and momentum of the electrons. For highly relativistic electrons, we have $E \propto |\vec{p}|$ instead; and going through the same calculations (neglecting the electron mass entirely), we find that the total energy of a relativistic fermion gas is proportional to $N^{4/3}/V^{1/3} \propto M^{4/3}/R$.

The gravitational binding energy, on the other hand, remains negative and proportional to $M^2/R$. This implies that the overall energy is itself proportional to $1/R$, and there is no extremum of the total energy of the system. Since the fermion energy and the binding energy always increase or decrease at exactly the same rate, there will be no stable equilibrium radius. The star will either blow itself apart or collapse in on itself, depending on whether the kinetic energy of the fermions or the gravitational binding energy wins out.

$\endgroup$
10
  • $\begingroup$ most excellent explanation. Yours is probably the best treatment of the energy balance argument I have yet read, thanks for posting it- Niels $\endgroup$ May 24, 2021 at 17:03
  • 1
    $\begingroup$ I feel like this answer is missing something about the conditions at which electron capture by protons becomes energetically (and/or entropically) favorable. $\endgroup$
    – zwol
    May 25, 2021 at 15:21
  • 1
    $\begingroup$ @zwol: that's beyond my expertise, I'm afraid. Feel free to write an answer that explains the issue and I'll be happy to upvote it. $\endgroup$ May 25, 2021 at 15:44
  • 1
    $\begingroup$ @DescheleSchilder I'm not sure that the average distance being low means that the uncertainty in position has to be low. On average they're very close, but it's not like the adjacent electrons form a box which constrains the other electron's position $\endgroup$
    – llama
    May 25, 2021 at 16:50
  • 2
    $\begingroup$ @mithusengupta123: The original question only asks about white dwarves, not neutron stars, so this answer just discusses why there must be a maximum mass for white dwarves. You can make similar arguments for neutrons; that's how Tolman, Oppenheimer, & Volkoff originally figured out that there must be a mass limit for neutron stars. But the simple argument I gave neglects interactions between the particles other than their gravitational binding energy. This is an OK simplification for white dwarves but it fails pretty strongly for neutron stars. $\endgroup$ Sep 13, 2021 at 11:55
6
$\begingroup$

An alternative: As the white dwarf mass increases, the electrons become ultra-relativistic. Hydrostatic equilibrium is not possible for ultra-relativistic degeneracy pressure.

Hydrostatic equilibrium requires: $$\frac{dP}{dr} = - \rho g \ . $$ Working just with proportionalities, non-relativistic degeneracy pressure $\propto \rho^{5/3} \propto M^{5/3} R^{-5}$, where $\rho$ is density, $M$ is mass and $R$ is radius. So the LHS and RHS of the hydrostatic equilibrium equation can be written \begin{eqnarray} M^{5/3} R^{-6} & \propto & (M R^{-3})(M R^{-2}) \\ & \propto & M^2 R^{-5}\ . \end{eqnarray} For a given mass, the radius can be adjusted to find an equilibrium.

For a more massive star, that equilibrium radius goes as $R \propto M^{-1/3}$, so a more massive star has a smaller radius, higher density; the electron Fermi energy increases and the electrons become ultra-relativistic.

Ultra-relativistic electron degeneracy pressure is proportional to $\rho^{4/3} \propto M^{4/3} R^{-4}$. Inserting this into the hydrostatic equilibrium equation we see \begin{eqnarray} M^{4/3} R^{-5} & \propto & (M R^{-3})(M R^{-2}) \\ & \propto & M^2 R^{-5}\ , \end{eqnarray} and thus there is no possible adjustment in radius that can make this equation balance. It is satisfied (but unstable) for just one mass - the Chandrasekhar mass.

$\endgroup$
0

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.