CarlB said:
It turns out that the Cambridge and Schwarzschild solutions use the same coordinates, other than coordinate time.
So do Painleve and Schwarzschild coordinates. Basically a certain time transformation maps one into the other. Time and space are not at right angles in the Painleve-Gullstrand coordinate system however, something you were complaining about earlier. But the PG coordinates are finite at the event horizon.
Take a look at, for instance,
https://www.physicsforums.com/showthread.php?t=126307
Unfortunately I have more work to do to sort out all the sign issues with the lograthmic conversion of the arctanh() function, they drive Maple nuts :-(.
[add]OK, I think this is all sorted out now, and it seems to match up well with the online homework solution
http://www.physics.umd.edu/grt/taj/776b/hw1soln.pdf
So the orbits are identical. Where the GR particles get stuck on the event horizon, the gauge particles go through to the singularity.
I think from this remark that you are taking the coordinates too seriously. A change of coordinates never affects the physics. The particles are still stuck on the horizon from the POV of the obsever at infinity, and the particles were NEVER stuck on the horizon from their own POV.
When we change the coordiantes, we do not change the particles orbit at all. We change only the description of the particle's orbit, we do not change anything physical.
An example might help.
Suppose we consider the coordinate system of an observer with a constant 1 light year/year^2 acceleration - a rindler observer, in a spaceship heading away from Earth with a constant acceleration.
The rindler observer, in his spaceship, obsevers an event horizon 1 light year behind him, the so-called rindler horizon. This is very similar to the event horizon of a black hole.
[add]
The metric for the accelerating spaceship , assumed to be accelerating in the z direction, is
-(1+gz)dt^2 + dx^2 + dy^2 + dz^2
You can see that at gz=-1, the cofficient of dt^2 becomes zero, just as it does in the Schwarzschild coordinate system. This marks the "event horizon" of the Rindler obsever, a plane located at z = -1.
[end add]
The rindler observer never sees the Earth as being more than 1 year old, because light emitted after that date never catches up with him as long as he continues to accelerate. (See anything on "hyperbolic motion" for why this is true, I think there is even something in Wiki about this).
The rindler observer then concludes (if he takes coordiantes too seriously) that the Earth never gets older than 1 year, and that the Earth never actually passes through the rindler horizon, 1 light year behind him.
The Earth observer does not observer anything unusual at 1 year, and cheerfully passes through "the horizon" (which doesn't even exist in his coordinate system) with no difficulty whatsoever.
Thinking that the Earth never gets more than 1 year old because the rindler obsever in the acclerating spaceship doesn't see it happen is a lot like thinking that the particle never falls into the event horizon because the outside observer doesn't see it.
Basically, your simulation is about coordinates, about how quantities that are infinite in one coordinate system are finite in another. Gauge gravity doesn't have anything to do with the issue at all, and I think your approach is going to mislead poor newbies.
It's basically standard GR, but with the coordinate system chosen to be flat. If you wanted to choose some other coordinate system you could do it that way too.
A picky point here. Just changing the coordinates is not enough to change curved space-time into flat. (Coordinates never affect the physics! Coordinates are just labels on a map, and the map is not the territory.) You have to go further, and change the metric as well, i.e. the defintion of length and time.