Gravitational Equivalent of Rindler Space

From this I conclude that the pseudo-gravity associated with an accelerated frame of reference in free space is not the result of curvature, but the true gravity in the vicinity of massive objects is the result of curvature. In this respect, the two situations are fundamentally different. Now I have to go back and think a little more about free fall in which, somehow, the curvature vanishes (at least locally).

Yes, they are fundamentally different. But just to throw in another example, the Langevin frame in SR represents a rotating observer with an inward radial acceleration. Written in the Born chart this frame has non-trivial curvature! One component of the Riemann tensor is non-zero. This excellent article has the details. One needs to understand frame fields in GR to fully appreciate it, but well worth a read.

http://en.wikipedia.org/wiki/Born_coordinates

Note, it is not the 4-manifold that has curvature; that is impossible. It is quotient 3-manifold that is Riemanian (not semi-Riemannian) that has curvature. This is no more surprising than that curved 2-manifolds can be embedded in a flat 3-manifold.

Thanks for the clarification. It would indeed be strange otherwise. Question - does this mean that the observer sees spatial curvature ? I suppose the spatial projection tensor hold the answer.

Yes, it implies that observers riding on a rotating disk, taking only local measurements, will detect non-Euclidean spatial geomoetry.

One stumbling block I have encountered has been in trying to get an understanding of the relationship between accelerated reference frames in free space, and reference frames at rest in the vicinity of gravitationally massive bodies, within the framework of the equivalence principle. Somehow I got it in my head that, from the perspective of observers in an accelerated reference frame in free space, their space is curved.

What we call the "curvature of spacetime" has a technical meaning; the equations that describe it are very similar to the equations that describe, say, the curvature of the earth's surface in terms of latitude and longitude coordinates, or any other pair of coordinates you might choose. This "curvature" need not manifest itself as a physical curve "in space".

For the rest of this post let's restrict our attention to 2D spacetime, i.e. 1 space dimension and 1 time dimension, i.e. motion along a straight line. This is adequate for many scenarios, even for black holes provided we are interested only in particles moving radially.

In the absence of gravitation, an inertial frame corresponds to a flat sheet of graph paper with a square grid. If we switch to a different inertial frame we "rotate" to a different square grid, but it is the same flat sheet of paper. (The words "rotation" and "square" here are relative to the Minkowski geometry of spacetime, which doesn't look quite like rotation to our Euclidean eyes, but nevertheless it preserves the Minkowski equivalents of "length" (spacetime interval) and "angle" (rapidity).)

If we switch to a non-inertial frame (but still in the absence of gravitation), we are now drawing a curved grid, but still on the same flat sheet of paper. Thus, relative to a non-inertial observer, an inertial object seems to follow a curved trajectory through spacetime, but this is due to the curvature of the grid lines, not the curvature of the paper which is still flat.

When we introduce gravitation, the paper itself becomes curved. (I am talking now of the sort of curvature that cannot be "flattened" without distortion. The curvature of a cylinder or cone doesn't count as "curvature" in this sense.) Now we find that it is impossible to draw a square grid to cover the whole of the curved surface. The best we can do is draw a grid that is approximately square over a small region, but which is forced to either curve or stretch or squash at larger distances. This grid defines a local inertial frame, where it is square, but that same frame cannot be inertial across the whole of spacetime.

So, to summarise, "spacetime curvature" refers to the curvature of the graph paper, regardless of observer, whereas visible curvature in space is related to the distorted, non-square grid lines drawn on the graph paper, and depends on the choice of observer.

Thanks very much DrGreg. You summarized the situation very well. Except for my conceptual "glitch" of failing to recognize that Rindler space is flat, I had already envisioned the remainder of what you so concisely and simply presented.
As someone studying relativity on my own, it has been hard to proceed without having someone to answer my questions. Discovering Physics Forums has been a real "find." I appreciate everyone's help.
Now for one additional question.
Since accelerated frames of reference are basically flat, and, in this respect, differ from reference frames at rest with respect to massive bodies, would it be at least conceptually possible for residents of an accelerated reference frame to experimentally measure the metrical properties of their space, and by doing so, ascertain that they are not experiencing gravitational curvature of spacetime?

This might be relevant:
http://www.mathpages.com/home/kmath530/kmath530.htm