exchemist:

What I still can't work out, though, is how he thinks the mass of a black hole (a concept that did not exist in his day) exerts its gravitational influence, if the space inside is discarded from any physical theory.

That's actually a valid question even when the space inside is

*not* discarded from the theory.

Remember that the GR description of gravity involves a curving of spacetime. The space around

*any* spherical mass is curved in approximately the same way. The same GR solution applies to the gravity of the Earth or the Sun, for instance, as applies to a black hole.

Mass causes the space around it to curve, which from a distance makes things look as if a force is acting to attract other masses. But gravity isn't a force in the GR description.

The curvature of spacetime around a black hole is caused by the mass of the hole. That doesn't require anything to propagate from inside the hole to the outside. You can imagine the "fabric" of spacetime stretching as more mass is added, like putting a heavy bowling ball on the surface of a trampoline then gradually adding more weight. Even the "space" far from the ball curves more as weight is added. Of course, this is an imperfect analogy.

The gravitational "field" of a black hole is sometimes referred to as a "fossil" or "vestigal" field. Whatever mass caused spacetime outside the hole to curve is no longer visible from outside and it can no longer send any kind of message out from the inside. So, in a sense, it doesn't matter whether the mass is there or not, any more. (Having said that, it appears from Hawking radiation etc. that the total mass-energy of a black hole is conserved, so it looks like whatever mass there is doesn't "go" anywhere inside the hole.)

It's interesting to think about how a black hole can possibly grow as more matter falls into it. I think a good explanation is that as a piece of matter falls in, the event horizon in effect expands

*outwards* towards the infalling matter - just enough to exactly account for the increase of mass related to the infalling matter and simultaneously shielding that matter from view.

The time comment is interesting, though, in the context of the present discussion. Does this start to make sense of Mike's conjecture? If in-falling mass takes an infinite time to collapse inward, to outside observers, can that mean it** looks to us** as if it is stuck at the edge of the event horizon?

It would

*look* like that, if we could see it. But light emitted outwards from the hole gets more and more Doppler shifted as it comes from closer and closer to the event horizon, until it is infinitely red-shifted at the horizon. So, even though, if we disregard red-shift, it would look from the outside as if all the infalling mass was "stopped" at the horizon, in fact the horizon looks black to us.

But see my post to Mike, above. This "outsider" view of the hole is just one choice of spacetime coordinates - the one we happen to have as observers located at a safe distance outside the hole. A free falling observer would not suddenly find himself surrounded by matter as he fell through the horizon. There is no force that would stop him or anything else falling past the horizon.

Why would the number of photons diminish?

Talking about light emitted from the hole again...

You can actually think of this in one of two ways: using a wave model of light, or a particle (photon) model.

In the wave model, the light waves are stretched out (red-shifted) as they propagate radially outwards. (Again, worth bearing in mind that the physics is the same whether we're talking about the Earth's gravity or a black hole's gravity.) As the waves stretch, their energy decreases.

In the particle model, the number of photons per second emitted by a regular source falling towards the horizon decreases continuously due to gravitational time dilation - time seems to run slower near the horizon than it does far from it. The photons themselves are also "stretched out" as they propagate outwards, again resulting in a red shift of the observed light.