Considering that space-time is expanding at an increasing rate or perhaps during times of inflation - could a black hole just above the neutron star threshold experience a decrease in density to switch back to a neutron star?

Edit: That is, same mass/energy as before but it's now contained in a volume greater than the Schwarzschild radius due to expansion (current or future or inflation era) in the fabric of space time itself.

PS: I'm not talking about that process by energy or mass escaping it (as discussed here).


2 Answers 2


No. According to accepted theory, a black hole generally can't lose energy or mass. Even with Hawking radiation, the black hole would simply change size accordingly and remain a black hole. Once it got too small it would evaporate into a burst of Hawking radiation.

I know you say you aren't talking about energy or mass escaping the black hole, but the idea of black holes is that they are a one-way process.

I fail to see how the linked article isn't an answer to your question.

  • $\begingroup$ same mass/energy but now contained in volume exceeding Schwarzschild radius due to space-time expansion $\endgroup$ Commented Feb 8, 2017 at 18:31

Spacetime does not expand at scales comparable to a Schwarzchild radius, or a solar system, or even a galaxy. Not trying to be a smarty-pants, but for the benefit of all, please get the basics under wraps before moving into conjecture.

  • 2
    $\begingroup$ For the benefit of all, can you be more specific about the rate mismatch? Expansion rates from the past. present or future predicted? $\endgroup$ Commented Feb 8, 2017 at 18:53
  • $\begingroup$ en.wikipedia.org/wiki/Metric_expansion_of_space $\endgroup$ Commented Feb 8, 2017 at 19:04
  • $\begingroup$ Thanks. Are you able to construct a cohesive answer? $\endgroup$ Commented Feb 8, 2017 at 19:06
  • $\begingroup$ The answer seems to contradict the big rip en.wikipedia.org/wiki/Big_Rip $\endgroup$
    – user126422
    Commented Feb 9, 2017 at 5:03

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.