The car on the other side of the RH is what cannot be represented in the Rindler frame of the accelerating car. It cannot be keeping up with the accelerating (but stationary) car in the ARF. (Please tell me if any of these acronyms are confusing. I tire of typing the full words). — noAxioms
The Rindler horizon's similarity to the event horizon is only insofar as any light travelling from the negative x direction cannot reach the x=0 worldline. That seems to be the entire basis for your argument that the space the photon travels in does not exist in the Rindler frame. This is exactly the same as saying the car behind does not exist in the frame of the car in front because it can never reach it. It's the same argument.
Yes, and the math says the acceleration cannot be finite at the RH — noAxioms
Except for the x=0 (light-like) worldline, yes. But...
which is why I said the length of my object extended almost 1 to the rear, but not all the way, because I wanted to avoid the infinite acceleration required there, with yes, infinite time dilation, just like at the EH of a black hole. — noAxioms
But you don't have to do this. If the acceleration of the leftmost part of the ship is finite at T=t=0, it will not follow the x=0 worldline (the light-line) but one of the x>0 ones. Infinite acceleration is the only way to follow the x=0 worldline. The leftmost end and rightmost end will not have the same horizon in the Minkowski frame, i.e. the horizon is proper-frame--dependent.
Two observers having the same proper acceleration (Bell's spaceships). They are not at rest in the same Rindler frame, and therefore have different Rindler horizons — https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox
(I am thinking of changing my username to BornRigid but I might get banned.)
As per Bell's paradox, there exists an inertial frame of reference in which we can arrange for the acceleration of the ship to be the same at both ends. As per Wiki:
Consider two identically constructed rockets at rest in an inertial frame S. Let them face the same direction and be situated one behind the other. If we suppose that at a prearranged time both rockets are simultaneously (with respect to S) fired up, then their velocities with respect to S are always equal throughout the remainder of the experiment (even though they are functions of time). This means, by definition, that with respect to S the distance between the two rockets does not change even when they speed up to relativistic velocities. — https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox
This ship would break apart because each section of it would be length contracted while maintaining its distance from the adjacent section, but the take-home of this is that it doesn't matter where the two ends are. One can always construct another inertial frame at rest wrt the first in which either end is zero or both non-zero. There is nothing special about the initial coordinate of each end of the ship because each end has its own Rindler horizon in its proper frame.
When we choose
one of the proper frames, the worldlines of each end will be constrained by the Rindler horizon of the proper frame we choose (either to stay to the right or to the left of it depending on whether the initial position of the
other end is to the left or the right of the light-line. Let's take the rightmost end as the proper frame. In its frame, it is not moving, and the left end is moving away from it. To see this, refer again to:
choose the point on the worldline to the right where t=1, and follow the t=1 line back to the other end of the ship. If the leftmost ship is still at x=0 (because it started at X=0), it is
still at T=0 in the Minkowski frame, and its velocity is therefore zero: no length contraction occurs at the leftmost end because it is not yet moving. (Length contraction is a function of
velocity, not acceleration.)
This ship would break apart in the proper frame of the rightmost end because parts to the left are moving away from the rightmost part with increasing acceleration. The leftmost part's coordinate still exists in the rightmost part's proper frame (indeed, we just traced to that coordinate with our finger).
As I said before, if the length of the ship in the proper frame of the rightmost part is infinite and the leftmost part is at X=x=0 at T=t=0 and the rightmost part has a finite X at some time T, then yes, in the Minkowski frame there will be infinite length contraction fitting the infinite length of the ship into the finite length between the x=0 worldline and X, because in the proper frame of the rightmost part of the ship (at infinity), the leftmost part of the ship is receding away at infinite acceleration. But since this is a physical impossibility, don't worry about it. For all finite lengths, the length contraction is zero or finite everywhere except at t=infinity.