Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

No physicist/astrophysicist I; All I know about the Chandrasekhar limit is that it apparently limits the mass a star may survive, beyond which it degenerates to a neutron star, or a black-hole.

Does something similar apply to a black-hole? Is there an upper limit to black-hole mass? Can it acquire infinite mass? Is there a limit beyond which it may no longer acquire mass? Does a black-hole die? Is there a corresponding time-frame applicable?

share|cite|improve this question

The Chandrasekhar limit is the maximum size of an object that can hold itself up against it's own gravity. A black hole, by definition, isn't held up - it has collapsed into nothingness.

The 'size' of the black hole isn't well defined, it depends on your theory, but in classical physics it's size would be zero - a singularity or infinite dense point.
The only size of a black hole you can talk about sensibly is the Schwarzschild radius which is the distance away from the center at which the escape velocity is the speed of light. Since nothing can go faster than light anything which passes this point is never coming out again. So in simple terms you can regard this as the boundary of the black hole experience.

As the mass of the black hole increases this distance increases. So large black holes can essentially grow for ever, and classically a non-expanding universe would end up with everything inside black holes. Small black holes can evaporate because of a quantum effect Hawking radiation. Smaller black holes evaporate faster - so there is actually a real small size limit.

share|cite|improve this answer
The size of a black hole is it's event horizon radius. The singularity is not a point in space where matter is scrunched up. This is something people say all the time, it's not right. The matter collapsed in a black hole has turned into its event horizon from the point of view of the outside observer, by merging with it. – Ron Maimon Dec 1 '12 at 21:58
I said in classical physics, in other physics it depends on what your model is. – Martin Beckett Dec 2 '12 at 1:30
It isn't really true classically either. The singularity is either to your future, or matter can't hit it (depending on whether it is spacelike or timelike). – Ron Maimon Dec 2 '12 at 2:39

First, I want to make clear precisely what the Chandrasekhar mass is: the maximum mass of a white dwarf supported purely by electron degeneracy. It depends on a few things (notably, the mean molecular weight of the white dwarf) and neglects other sources of or deviations to pressure, but its canonical value of 1.44 $M_\odot$ is a pretty accurate estimate at the mass above which a white dwarf or degenerate stellar core collapses.

The next possible source of support against gravity is neutron degeneracy. i.e. support from the fact that you can't squeeze neutrons into the same quantum state. The Tolman–Oppenheimer–Volkoff limit is the corresponding mass limit for the maximum mass of an object supported by neutron degeneracy. It is much more difficult to calculate this maximum mass because it requires precise knowledge of the equation of state and, currently, we just don't know exactly how matter behaves under those conditions. Even so, the limit is probably more than 2 $M_\odot$ because such a neutron star has been observed and broadly thought to be less than about 3 $M_\odot$. The upper end of the range is very rough, though.

Black holes, however, have no such mass limit because there is no pressure support. We are quite confident that there are black holes at the centres of distant galaxies with masses that exceed 10$^{10}$ $M_\odot$, and nearby M87 hosts a black hole of a few 10$^9$ $M_\odot$. Any apparent limit on the mass of a black hole is just because it hasn't been fed enough.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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