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I'm going to repeat an analogy I've made elsewhere...

Take out a pen and twirl it around. Its proper length is unchanged, but its apparent length depends on the perspective of your view. It exists in three-dimensional space, but is projected to your two-dimensional field of view, and this projection is what you see. The third dimension, "depth" is orthogonal to that two-dimensional view. Different observers will be oriented and hence have a different "depth" axis.

Now, what does it mean for different observers to have a different "orientation" in spacetime? Draw little arrows from the origin on the (t,x) plane--a vector [t;x] almost begs to be interpreted as a velocity: x distance in t time. It defines the observer's time axis, and analogously to depth, the observer's space is orthogonal to this axis. The pen exists boths in space and in time--it is inherently four-dimensional. The observer's view of an object is that object projected onto the space orthogonal to the velocity vector, and this projection is what the observer measures. Just as differently oriented observers in space have see different lengths for the same objects, differently oriented observers in spacetime (i.e., different velocities) as well.

So, to summarize the analogy: 2D field of view + 1D depth axis <--> 3D space + 1D time axis.

I think this interpretation is very intuitive. The only layer added to common-sense experience is that spacetime orientation is a velocity. I hope it clears up the reason why I was concerned with rotations in the first place.

This is farly hard to explain without a digram, but if you draw your own hyperbolas as you follow along, it shouldn't be too difficult.

Let's say, at t = 0, the object is some distance x = d away from the origin, initially at rest, thereafter undergoing a constant acceleration in the positive x-direction. A constant acceleration, as seen above, is simply a constant rotation. So what are the coordinates at time t? Rotating the axes one way is equivalent to rotating points the opposite way, so

(See the previous page for the same transformation without matrices.) This is the right branch of a hyperbola (cf. eqns. 19-20). The asymptotes of this hyperbola pass through the origin.

Now, think of how an observer at x = 0 could communicate with the object. Light signals sent in the positive x-direction at t_0 by the observer travel along the lines of slope 1, x = t - t_0 (or x = c(t - t_0) in standard units), corresponding to "one light-second per second" or however the units are normalized. Conversely, light signals sent by the object in the negative x-direction travel along lines of slopes -1. This means that the object will always be able to send signals to the observer, but the converse might not hold--if the asymptote of the hyperbola might be have slope 1 or below!

[Edit: This happens for in every case--there since sinh aT ~ cosh aT for large T, but the placement of the asymptote is important--what I intended to convey is that if the object starts out with distance greater than 1/a [c²/a], then it is always behind the horizon for t≥0.]

By rescaling your units in the equation of the hyperbola, you should be able to figure out that this happens when d≥1/a (or d≥c²/a in standard units). The acceleration horizon is the asymptote of the hyperbola corresponding to the wordline of the accelerated object. It's a one-way limit to communication.

It's fairly analogous to event horizons in GTR, actually--there, an object enclosed by a horizon can still receive information from outside the horizon but cannot send any to the outside. Here, the observer can receive information from the accelerated object but cannot send any. In both cases, the horizon is one-way. Acceleration horizons even emit Unruh radiation just as black hole horizons emit Hawking radiation, much by the same kind of effect.

(Apologies for not explaining it clearly enough previously.) Oh, that happens in every case (which makes sense, since constant acceleration in the object's frame should be asymptotic to lightspeed in the observer's frame)--what differs is the placement of the horizon. Here is the spacetime diagram the wordlines three objects (red) with their corresponding horizons (magenta). The blue lines represent light signals either outward (slope 1) or inward (slope -1). One of the objects starting at the origin has acceleration rate 1/4 of the other. As you can see, each object's wordline intersects every line for slope -1 for the t>0 region--i.e., it can always send signals to the observer at the origin. The converse is not true--the signals from the origin colored magenata will never catch up with the accelered object. Those asymptotes are the horizons for those objects.

Oh, it's no problem. STR is a lot easier from the spacetime perspective. With a proper diagram, many things suddenly become obvious.