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Comment by pdonis | Hacker News Reader

given the black hole mass large enough, the gravitational force at the Schwarzschild radius can be as small as we'd like it to be

No, it can't; the "acceleration due to gravity", which is the acceleration required to "hover" at a constant altitude above the horizon, diverges as the horizon is approached.

What can be made as small as desired by making the hole's mass large enough is tidal gravity at the horizon.

several billions years ago, when observable Universe had 10B radius, it would be a black hole

No, it wouldn't. A black hole is a stationary spacetime. The spacetime of the universe is not stationary. The spacetime model that describes the universe is very different from the spacetime model that describes a black hole; the fact that you can plug the mass of the observable universe into the Schwarzschild radius formula and get a number out does not mean that number has any physical meaning for the universe.

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>No, it wouldn't. A black hole is a stationary spacetime. The spacetime of the universe is not stationary. The spacetime model that describes the universe is very different from the spacetime model that describes a black hole

exactly. There is no stationary spacetime in our Universe. Black hole is pretty artificial model where pure mathematical artifacts of singularity at the horizon is taken for the real thing.

> the fact that you can plug the mass of the observable universe into the Schwarzschild radius formula and get a number out does not mean that number has any physical meaning for the universe.

Taking a big chunk of space and calculating escape speed from its gravitational field - how is that not a physical meaning? At what specific size of the chunk you think it becomes not meaningful?

There is no stationary spacetime in our Universe.

The universe as a whole is not stationary, nor even close to being so; but portions of the universe are very close to being stationary. The solar system, for example. Black holes do not have to be exactly stationary; if they are as close to being stationary as the solar system, that's plenty close enough.

Black hole is pretty artificial model

The exactly spherically symmetric solution is an idealization, yes; but there are plenty of numerical simulations that show that non-symmetric spacetimes still form event horizons.

where pure mathematical artifacts of singularity at the horizon is taken for the real thing.

There is no physical singularity at the horizon. Some coordinate charts have a coordinate singularity at the horizon, but that's easily fixed by just using a different coordinate chart. The only physical singularity is at r = 0.

Taking a big chunk of space and calculating escape speed from its gravitational field

Escape to where? You can't escape from the universe as a whole. The concept of "escape speed" has no meaning for the universe as a whole.

>Escape to where? You can't escape from the universe as a whole. The concept of "escape speed" has no meaning for the universe as a whole.

take a 1B light years radius ball, populate it with density of our Milky Way - that ball will have 1B Schwarzschild radius. Calculation of gravitational potential (escape speed) from it to the rest of the Universe makes sense, doesn't it?. And due to this gravitational potential it will be bona fide black hole from the point of view of the rest of the Universe.

Calculation of gravitational potential (escape speed) from it to the rest of the Universe makes sense, doesn't it?

Yes, but that's not the same as escaping from the universe as a whole.

And due to this gravitational potential it will be bona fide black hole from the point of view of the rest of the Universe.

So what? What does that have to do with assigning a gravitational potential to the universe as a whole?

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