[Mike Gale (OP)]

It is well known that the speed of light goes to zero at the event horizon of a Schwarzschild black hole. You can prove this to yourself by solving the metric for a radially in-falling photon, for which proper time is identical to zero. The result is:
c=c_o(1-r_s/r)
The Shapiro effect exemplifies this phenomenon.
What happens to an in-falling mass is less obvious because the speed of proper time is a function of velocity:
dT/dt=sqrt((1-r_s/r)(1-(v/c)²))

[evan_au]

As I understand it, what happens depends on the frame of reference of the observer:
- If you are an observer on the falling particle, you cross the event horizon in the blink of an eye.
- if you are an observer distant from the black hole, you never see the particle crossing the event horizon.

It is what happens after the particle crosses the event horizon that is more mysterious.

[jeffreyH]

The freely falling frame is inertial since it is not supported against gravity. No force is felt. Proper time must continue up to and beyond the horizon. Beyond the horizon our definition of proper time may or may not be correct.

[Mike Gale (OP)]

One expert I consulted on this issue had this to say:
1)  The term 'proper time speed' is unconventional and may lead to confusion. The conventional term is 'time dilation'.
2) The expression I quoted for that quantity is obtained by dividing both sides of the metric by 'dt'. That's invalid if dt=0, which may be the case in some contexts.
3) The metric permits trajectories that are not geodesic. Although a geodesic describes a trajectory that is consistent with the metric, the reverse is not necessarily true.

[Mike Gale (OP)]

Even so, this formulation of the metric suggests that dT and dr go to zero at the horizon:
(c_o*dT)²=((c*dt)²-dr²)/(1-r_s/r)-(r*d(angle))²
The question is whether there is a geodesic for which that is the case and, if so, does it approximate Newtonian free fall in the weak field limit?

[Mike Gale (OP)]

Another way to look at it is:
(c_o*dT)²=(c*dt)²*(1-(v/c)²)/(1-r_s/r)-(r*d(angle))²
If d(angle)=0 and (v/c)²=r_s/r then:
dT=(1-r_s/r)dt
Is that a geodesic? If so, then v=0 at the horizon.

[Mike Gale (OP)]

Wikipedia has a good article on Schwarzschild geodesics: https://en.wikipedia.org/wiki/Schwarzschild_geodesics
In a nutshell, evan_au is correct. Penetration is a mundane occurrence for a local observer. However, the local perspective is inconsequential for the rest of us. From our point of view, the black hole cannot accumulate mass. Furthermore, although Wikipedia gives exact solutions for time dilation in the free fall case, it does not address spatial dilation. Flamm's paraboloid teaches us that radial distances are dilated in the local reference frame. The gravitating mass recedes due to spatial dilation so penetration may well be impossible in any context.
In any case, the distant observer perceives objects to accelerate towards the gravitating mass in a weak field context and stop dead in their tracks at the horizon so there has to be a point in between where they start to decelerate. You can't observe an object impacting the horizon because it is red-shifted into oblivion and eternally hidden by the Shapiro effect. But you can extrapolate its trajectory to predict the speed of impact. Conventional wisdom says impact velocity is finite. I disagree, but I can't find a mathematically vigorous argument to make or break my case. The best I can do is appeal to causality. A reference frame in which mass can out pace light doesn't make any sense.

[Mike Gale (OP)]

The weird thing about SC time keeping is that coordinate time and proper time are both functions of coordinate radius. Coordinate time is a bookkeeping concept. Nobody actually ages at that rate. It's more like an optical illusion.

[Mike Gale (OP)]

I think this is what I'm talking about: https://en.wikipedia.org/wiki/Gullstrand–Painlevé_coordinates

The article confirms the expression I derived for coordinate velocity in the radial free fall case. The in-falling object does indeed appear to decelerate and stall at the horizon. They call this result an "accounting entry" since coordinate velocity is reckoned by the elapsed time between messages, which are delayed by the Shapiro effect. It is an illusion, but a stubbornly persistent one because the horizon is inarguably impenetrable from that perspective.

[Mike Gale (OP)]

The question is, does the raindrop affect the event horizon in coordinate space when it crosses over in the local reference frame? I think the answer is no because changes in the field must not propagate faster than light.

[PmbPhy]

It is well known that the speed of light goes to zero at the event horizon of a Schwarzschild black hole. You can prove this to yourself by solving the metric for a radially in-falling photon, for which proper time is identical to zero. The result is:
c=c_o(1-r_s/r)
The Shapiro effect exemplifies this phenomenon.
What happens to an in-falling mass is less obvious because the speed of proper time is a function of velocity:
dT/dt=sqrt((1-r_s/r)(1-(v/c)²))

While there's a question in the subject line, the answer of which depends on the observer, there's no question in this opening post. What is the question here?

[evan_au]

From our point of view (as distant observers), the black hole cannot accumulate mass.

Due to Newton's shell theorem, any mass closely approaching the black hole's event horizon appears (from the viewpoint of a distant observer) as if the black hole has gained that mass, despite the fact that the observer never sees it actually cross the event horizon.

[Mike Gale (OP)]

Accumulation of mass outside the horizon is not the same as crossing the horizon.

It is well known that the speed of light goes to zero at the event horizon of a Schwarzschild black hole. You can prove this to yourself by solving the metric for a radially in-falling photon, for which proper time is identical to zero. The result is:
c=c_o(1-r_s/r)
The Shapiro effect exemplifies this phenomenon.
What happens to an in-falling mass is less obvious because the speed of proper time is a function of velocity:
dT/dt=sqrt((1-r_s/r)(1-(v/c)²))

While there's a question in the subject line, the answer of which depends on the observer, there's no question in this opening post. What is the question here?

I'm trying to figure out how a Schwarzschild black hole grows from the perspective of the bookkeeper, who is infinitely removed from the central singularity. From that perspective, it takes an infinite amount of time for anything (including light) to reach the horizon so it is, in effect, impenetrable and there is no opportunity for growth. Everything piles up just outside. If the pile up is radially symmetric, its centre of mass will be coincident with the central singularity so the bookkeeper will perceive that as growth. A hollow shell of mass looks the same as a central singularity when viewed from the outside so it would seem that the latter is a fiction.

[Mike Gale (OP)]

From our point of view (as distant observers), the black hole cannot accumulate mass.

Due to Newton's shell theorem, any mass closely approaching the black hole's event horizon appears (from the viewpoint of a distant observer) as if the black hole has gained that mass, despite the fact that the observer never sees it actually cross the event horizon.

I think you meant to say Gauss's theorem (of divergence.) Newton's shell theorem addresses the field strength inside the boundary. But I see your point. A sufficiently dense shell of mass will form a black hole and it is entirely possible that the masses are pushed together (e.g. by a collapsing star) rather than being pulled together by a pre-existing singularity. The primordial case would be a perfectly head-on collision between two elemental masses.

[Mike Gale (OP)]

Negative mass puts an interesting twist on the primordial case because:
a) Positive mass attracts negative mass.
b) Negative mass repels positive mass.
There is no net force on either particle as they approach one another so they can get arbitrarily close and annihilate (or rather nullify.) I don't know if that leaves a singularity or just flat space.

[Bill S]

I've just met this assertion by Christoph Galfard (The Universe in your Hand); and am trying to process it in terms of SR.

  He says of an asteroid that appears to be stationary on the event horizon of a black hole:

 ”It is gone………..Only its image remains. That is the distortion of spacetime in action. ,,,,,,,,Our time, yours and mine, is not ticking like the rock’s.  The asteroid is beyond the horizon.  Its image is still on the horizon.  That’s how it is.”

[Mike Gale (OP)]

An image is an apt analogy because the asteroid has no radial extent at the horizon, but the image is undetectable due to infinite redshift. I find the latitude/longitude analogy more instructive. It's like the in-falling object is heading towards a spacetime cliff and it takes an infinite amount of time to reach the brink, where it can no longer make headway in the radial direction. The horizon (at R=2m) is the edge of the perceptible universe in the same sense as R=infinity.

[jeffreyH]

The object crosses the horizon. The light reflecting off it is slowed by the gravitational field and takes longer to get to the remote observer situated further from the horizon. It appears to slow down.

[Mike Gale (OP)]

An object can cross the horizon in a finite amount of proper time, but the cross-over event occurs at the end of coordinate time. The same is true of light so you can't illuminate objects on the horizon. Kevin Brown breaks it down here: http://mathpages.com/rr/s6-04/6-04.htm, He provides a nice visualization, too.

[jeffreyH]

Without an actual black hole to play with it is all conjecture.