I've not noticed before that special relativity implies a 'wall of death' as you put it but I see now you can get that from the Lorentz transform t'=(t-vx/c^2)/sqrt(1-v^2/c^2) combined with v=at.
Does the z=-c^2/a relation still hold in general relativity or is it modified by other terms?
Also
> we can't see beneath our feet
I presume you mean that we can, but for the values of A and z we experience here on earth, the radius of the Earth is much smaller than z so a point at a distance z "beneath our feet" doesn't exist as it's up and out the other side of the gravitational well
> And now we have a photo of a black hole to prove it!
I'm guessing you mean the event horizon is exactly such a "wall of death"?
Yes, the event horizon of a black hole is precisely this sort of wall of death. Indeed there is an active debate in the literature about whether outside observers ever see black holes because from the outside perspective infalling mass should become “frozen” on the surface, with Liu and Zhang very recently in 2009 arguing that if you consider a shell of nonzero thickness falling symmetrically into the black hole then as this mass approaches the event horizon the event horizon actually expands to gobble up some of the matter—this paper is then cited and refuted in a 2011 paper in the same journal by Penna, who argues that they got one little detail very wrong and that once you correct for this you do indeed see the matter “frozen” on the wall-of-death; see [1] for a PDF of this paper.
The z = -c² / A relation does hold even when you transfer from the first-order-in-β transform to the full Lorentz transform; for an outworking from the inimitable John Baez, see [2].
I did mean that we can’t see beneath our feet, but I see your point: at g = 10 N/kg this term c² / g is something like 10^16 meters away, way way outside of the Solar System, which is only in the billions of km large. I was purely thinking about our Schwarzschild radius, which is millimeters away from the center of the Earth: therefore we cannot see the Schwarzschild event horizon because it is nonexistent; it is “underground” but so far that then most of the mass is outside of that, so you have to recalculate and get an even smaller amount, but that then needs another recalculation... and so on. It vanishes to zero because the mass is not located in a condensed enough space.
I've not noticed before that special relativity implies a 'wall of death' as you put it but I see now you can get that from the Lorentz transform t'=(t-vx/c^2)/sqrt(1-v^2/c^2) combined with v=at.
Does the z=-c^2/a relation still hold in general relativity or is it modified by other terms?
Also
> we can't see beneath our feet
I presume you mean that we can, but for the values of A and z we experience here on earth, the radius of the Earth is much smaller than z so a point at a distance z "beneath our feet" doesn't exist as it's up and out the other side of the gravitational well
> And now we have a photo of a black hole to prove it!
I'm guessing you mean the event horizon is exactly such a "wall of death"?