In the wikipedia page "Future of an expanding universe" it refers to the scenario of a future without proton decay.

The page talks about how processes would lead to stellar-mass cold spheres of iron, calling these objects "iron stars":

In 101500 years, cold fusion occurring via quantum tunnelling should make the light nuclei in ordinary matter fuse into iron-56 nuclei. Fission and alpha-particle emission should make heavy nuclei also decay to iron, leaving stellar-mass objects as cold spheres of iron, called iron stars.

Under the heading "Collapse of iron star to black hole" it then says :

Quantum tunnelling should also turn large objects into black holes. Depending on the assumptions made, the time this takes to happen can be calculated as from 101026 years to 101076 years. Quantum tunnelling may also make iron stars collapse into neutron stars in around 101076 years.

How would quantum tunnelling lead to the collapse of an iron star to a black hole?

  • $\begingroup$ There's a pdf file at bit.ly/2aHylid if this does not connect you to the file, try searching for "BLACK HOLE PRODUCTION VIA QUANTUM TUNNELING by SN Solodukhin" it's on a cern server, so you could search their main site. $\endgroup$
    – user108787
    Commented Jul 26, 2016 at 14:31
  • $\begingroup$ @count_to_10 That paper is very interesting, but appears to deal with the creation of black holes from the collision of individual particles. What I don't understand is how quantum tunnelling could lead a stellar mass object such as the conjectured iron star to become a black hole over the stated time period. $\endgroup$ Commented Jul 26, 2016 at 15:27
  • $\begingroup$ Sorry, I just looked at the title and the first page of it. I have absolutely no idea how quantum tunneling would work, apart from a silly idea that the density would increase to BH levels if, eventually all the tunneled particles collected in a small enough region, a few at the start then, as more collect, the density increases. $\endgroup$
    – user108787
    Commented Jul 26, 2016 at 15:33
  • $\begingroup$ @count_to_10 No need to apologise, it's an interesting paper and may be related to the process I'm trying to find out more about. $\endgroup$ Commented Jul 26, 2016 at 16:19

3 Answers 3


First some context:

A cold iron "white dwarf" will be stable if it's mass is below about $1.2M_{\odot}$. This is the equivalent of the Chandrasekhar mass for a more normal white dwarf supported by electron degeneracy pressure, but is lower because there are fewer electrons per mass unit in a gas of iron.

However, lower energy configurations are possible - i.e. neutron stars and black holes if the electron degeneracy can be compromised.

One way of doing this is electron capture (a.k.a. neutronisation or inverse beta decay). This is where a proton in the iron nucleus captures an electron from the degenerate gas and turns into a neutron.

The process is "endothermic", the electron requires an energy of around 5 MeV. This can be achieved in a degenerate electron gas if the electron density is high enough to push the Fermi energy beyond this threshold.

If you do the sums, these densities are also reached just a little below $1.2 M_{\odot}$ (recall that more massive white dwarfs get smaller).

This process occurs in the crusts of neutron stars and builds up increasingly neutron-rich nuclei (note that iron is only the equilibrium nucleus in low density matter). However, it is not possible to build an entire star out of material behaving in this way because the removal of free electrons softens the equation of state dramatically and leads to collapse.

Now the answer:

In an iron white dwarf with mass below $1.2 M_{\odot}$ we would usually assume stability because the electron Fermi energy is below the neutronisation threshold. However, quantum tunneling will mean that occasionally, an electron capture will occur, leading to a more neutron-rich nucleus and one fewer electron in the gas. This lowers the degeneracy pressure, the star gets smaller to compensate and the electron density increases again. Repeat many times and eventually the material attains too few electrons per unit mass to support itself and it collapses.

In other words the tunneling gradually makes more neutron-rich material and lowers the Chandrasekhar mass until instability is triggered.

The result would be a neutron star.

I don't understand at all what the process to form black holes could be. Any long-lived, cold iron star must be less massive than $1.2 M_{\odot}$ as explained above. The collapse of such an object will always lead to a neutron star because this is well below the maximum mass that can be supported by dense neutron star matter.

Dyson's arguments in the paper referred to are very abstract (and unconvincing to me - for example he says there is no stable state for stars greater than the Chandrasekhar mass, although the first neutron star mass measurements, showing that they could exist with greater masses were not available in 1978). It may be possible to randomly form tiny black holes that then end up consuming the entire star. On the other hand, it is widely thought that tiny black holes will immediately evaporate in a burst of radiation. So this process seems unlikely to turn the whole star into a black hole and instead just gradually convert the rest mass into Hawking radiation.

Dyson does not directly mention quantum tunneling as anything that facilitates this process, though given that nucleons are repulsive at short range it would likely be required. The range of suggested timescales in Dyson's paper are connected with how small the smallest black hole can be - whether it is the Planck mass or the somewhat higher minimum mass for which a classical black hole description is meaningful and hence on the number of nucleons that have to be crammed into their Schwarzschild radius in order for this to happen.

  • $\begingroup$ A superb answer! Thank you so much. $\endgroup$ Commented Jun 15, 2020 at 8:06
  • $\begingroup$ Quantum tunnelling to a black hole would be the same way - perhaps tunnelling of quarks to very rare but denser states if any exist, or even directly to small black holes. However the latter case would far more likely just result in evaporation of the neutron star instead because the tiny blackhole would instantly evaporate. Since we don't know what physics fills the energy gap from about $10$ to $10^{16}\ \mathrm{TeV}$, we should probably not put too much on that Wiki prediction I'd say. $\endgroup$ Commented Oct 15, 2021 at 23:45

Extreme quantum tunneling

Look at the number we have: 101076. There is a double exponential. This number is huge, but what does it mean?

Quantum tunneling takes exponentially longer with higher and higher potential wells (or farther and farther distances). Roughly speaking, the time taken to cross a barrier of distance $d$ and height $E$ is proportional to $exp(d \frac {\sqrt {E m}} {\hbar \sqrt {2}})$, where $m$ is the mass of a particle. This means that, for a 1eV barrier, each 852 femtometers will approximately halve the probability if the particle is an iron nucleus.

Conversely, the "size" of the barrier in units of $d \frac {\sqrt {E m}} {\hbar \sqrt {2}}$ is roughly the natural log of the time taken. Which is $2.3\times 10^{76}$ (it does not matter if the original number is in years or zepto-seconds). Such a number is still huge even on a macroscopic scale.

This is about enough for $10^{46}$ nuclei (1% the mass of the moon) to simultaneously overcome a barrier $6000m$ wide (the radius of a sphere enclosing about this many nuclei at white-dwarf densities) and the Planck energy of $10^{27} eV$ high, as they all crush together into a small black hole which can then eat the white dwarf.

Doubly exponentially large numbers are contrary to our intuition about what cannot happen. In this case, "quantum mechanical tunneling can't happen at large scales". We also see similar effects with thermodynamics and monkeys on a typewriter. But our intuition is right: this will never be observed. Even in the unlikely event of protons not decaying, all sources of energy will run dry long before we can see it. Even mechanisms that "passively wait until it happens" are easily defeated by such huge numbers.

  • $\begingroup$ @PM2Ring Added explanation of m. $\endgroup$ Commented Oct 16, 2021 at 0:11
  • $\begingroup$ FWIW, there are various interesting points raised on physics.stackexchange.com/q/606390/123208 which also concerns an extremely unlikely event. One of the links there is to Probing the Improbable: Methodological Challenges for Risks with Low Probabilities and High Stakes: "If the probability estimate given by an argument is dwarfed by the chance that the argument itself is flawed, then the estimate is suspect". $\endgroup$
    – PM 2Ring
    Commented Oct 16, 2021 at 1:18
  • 1
    $\begingroup$ @PM: "Probing the improbable" is mostly about Black swan events such as the COVID outbreak. Events which are rare but make up with their rarity with high-impact (good or bad) and thus make statistical analysis of average risks extremally difficult. In contrast, this example is so vanishingly rare it is frank impossible for anyone to see. $\endgroup$ Commented Dec 20, 2021 at 16:21
  • $\begingroup$ Sure. I just wanted to mention that we should be cautious with probability estimates that may be smaller than the probability that the model itself is flawed. $\endgroup$
    – PM 2Ring
    Commented Dec 20, 2021 at 19:57
  • $\begingroup$ @PM: that is a good point. Conceptually, one easy way to incorporate this in most models is to add a small T-distribution with 1-3 DOF to the assumptions of the noise distribution. This allows for a small chance of human-error. Too bad it makes the math uglier (numerical gradient descent rather than simply a matrix inversion, medians instead of means, etc) because that acts as a honeypot to trap people. $\endgroup$ Commented Dec 21, 2021 at 22:19

Well iron stars will collapse due to quantum tunneling. Iron from the surface of the iron star over a really really really really really long time will go to the core. This will happen to all the iron atoms. Then the iron star will be so dense that it collapses into a neutron star. This neutron star then has the ability to turn into a black hole and by hawking radiation evaporate.

So in summary quantum tunneling tunnels iron nuclei from the surface of the iron star to the core forcing the density and gravity to go up until it turns into a neutron star and inevitably to a black hole.


  • $\begingroup$ Why would the Chandrasekhar limit be different for iron stars than it is for helium stars. Both would be supported by electron degeneracy pressure. $\endgroup$ Commented May 9, 2020 at 0:29
  • $\begingroup$ Both would be held up by degeneracy pressure but iron stars are denser than helium stars(so called white dwarfs). Also because eventually quantum tunneling will get around the degeneracy pressure. This process is extremely slow though. The process where it turns into a neutron star(then a black hole) is extremely slow. It will be 10^1500 years long even longer. $\endgroup$ Commented May 9, 2020 at 13:43
  • $\begingroup$ The only known way this could happen is if quantum tunneling turns a proton+electron into a neutron. I don’t see this process discussed in any discussion on iron stars. $\endgroup$ Commented May 9, 2020 at 15:23
  • $\begingroup$ Further, free neutrons decay with a half life of 10 minutes. So it’s clear an iron star would never become a neutron star even after 10^1500 years. $\endgroup$ Commented May 9, 2020 at 19:34
  • 1
    $\begingroup$ @RoghanArun that link does not say HOW it happens, it merely states that it happens with no explanation. $\endgroup$ Commented Jun 15, 2020 at 10:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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