Farsight:
I'm not mixing anything up at all. Einstein said a clock kept at this place would go at the rate zero.
Yes yes. But who is measuring the rate? Different observers measure different rates for the same clock, depending on their state of relative motion and their location in (curved) spacetime.
In this case, the rate zero thing refers to the point of view of a distance observer, far from the event horizon. The local observer, next to the clock, sees its proper time, which never stops.
Not true. Curved spacetime is associated with Riemann curvature, see the depiction on
wiki. That's associated with the tidal force.
Tidal force is a gravitational force (not that "force" is the correct term to use in general relativity, but you know what I mean, I hope).
There's no detectable tidal force in the room you're in.
Not detectable, perhaps, but it's there. It really depends on how you're doing the detection. The Earth's gravitational field is modelled using the Schwarzschild geometry, same as the black hole's. (Again, there's a caveat here about the Earth's rotation, but let's keep things simple.)
But lift up a pencil, and let go: gravity is readily detectable. Curved spacetime is the second derivative of gravitational potential, see the plot
here. You need the curvature to get your plot off the flat and level, but the local force of gravity relates to the local slope of the plot. Not how curved it is. For a very big black hole there's no discernible tidal force and no discernible spacetime curvature.
I'm not clear on what point you're trying to make. Are you claiming that there's no gravity at the event horizon? That is, if I placed a massive object at the horizon you say it wouldn't fall into the black hole? I think that's what you're saying.
Is it possible for a black hole to form, then, in your opinion?
James, with respect, the waterfall analogy is misleading, it's wrong, it's pseudoscience junk peddled by celebrity quacks like Max Tegmark.
Ok. It's not worth arguing. Forget the analogy. Let's just deal with actual situation as described by general relativity. Do you have any comments on the relative speed of the horizon etc., which I pointed out in my previous post? There's no analogy there; it's what GR tells you when you do the calculation.
But the coordinate speed of light is not. And the coordinate speed of light is important when we look to that favourite thing in relativity: the light clock.
At the event horizon of a black hole, Schwarzschild coordinates are next to useless. They don't describe the proper time or the perception of space that an observer has at the horizon. The Schwarzschild t coordinate tells us something about time as measured on a very distance clock sitting out in essentially flat space a long way from the horizon, but it certainly doesn't say anything about how clocks are ticking at the horizon itself.
Imagine you place light clocks in an equatorial slice around the Earth. Now go and look at that depiction on the wiki Riemann curvature tensor page.
What am I supposed to notice when I do that?
I'm not making any errors. You are.
Your claim that the proper time of an infalling observer stops at the event horizon is a clear error.
