Q-reeus said:
Thanks for the clarification - had a feeling you may have meant it that way. So I have no problem with that position; just not sure what advantage it brings to the table.
It makes reasoning easier, because you can use SR directly in a local inertial frame, so you can reason about what's happening locally without having to worry about the effects of curvature.
For example, take your scenario of a mass m at height h1 suspended by a cable which is attached to a motor at a higher height h2. The motor pulls the cable up by some small distance; what force does it have to exert, and how much work does it do?
First, we need to make sure that the problem is completely specified. To do that, we need to specify how the displacements of the cable are related at h1 and h2. Your previous statement of the scenario appeared, to me, to assume that the actual, physical displacements were the same; i.e., that the cable's motion was "rigid". We'll see what that means mathematically below.
Second, we need to clarify what it is we are actually trying to analyze. We are *not* analyzing the dynamics of the movement: the tension in the cable, the time it takes to move, time delay between the motor starting the cable moving at its end and the mass actually moving at its end, etc. We are only comparing two static equilibrium configurations: the "before" configuration, with mass at h1, and the "after" configuration, with mass at h1 + some small distance, and some additional cable retracted at h2. We are essentially assuming that the time it takes to transition from one static equilibrium to the other is very fast compared to everything else in the problem. This seems reasonable if the black hole is large enough compared to the difference in heights.
Given the above, we can write down a simple "work equation" in the local inertial frame of the mass m at height h1, centered around the event at which the displacement takes place:
dW_{1} = F_{1} dx_{1}
where dW1 is the work done on the mass m, F1 is the force applied, and dx1 is the displacement. Note that dx1 is a differential of the x coordinate in the local inertial frame, which we are assuming is pointed radially outward; and since it's a local inertial frame, dx1 directly represents the actual, physical displacement (because all the metric coefficients in a local inertial frame are 1--there is no "distortion" due to curvature).
We can also write down a simple work equation for the motor at height h2 when it retracts a length of cable:
dW_{2} = F_{2} dx_{2}
where dW2, F2, and dx2 are defined similarly.
The "rigidity" condition above let's us write:
dx_{1} = dx_{2} = dx
That is, the actual physical displacements are the same. So we can rearrange the two equations above to both be equations for dx, and set them equal:
\frac{dW_{1}}{F_{1}} = \frac{dW_{2}}{F_{2}}
Or, rearranging to obtain the quantity we want:
dW_{2} = \frac{F_{2}}{F_{1}} dW_{1}
That is, the work done by the motor is "redshifted" relative to the work done on the mass, by the same factor as the force is "redshifted".
Note that I didn't have to say *anything* about how the displacements in the local inertial frames related to coordinate differentials dr1, dr2 in the global Schwarzschild coordinates. That's simply irrelevant, given that you have already specified that the actual, physical displacements are the same. The local inertial frame analysis gives the simple answer directly, without having to deal with any such complications.
I'll refrain from commenting yet on the rest of your post, since it kind of seems like a work in progress...
