Q-reeus said:
It drops by a typically small fractional value (depending naturally on the shell thickness) in a smooth way.
This much is fine; but the relationship of the radial and tangential metric components to the potential is what I was getting at. See below.
Q-reeus said:
Why would there be - there is explicitly zero dependence on potential everywhere exterior to the shell according to SC's. And there is some strange physical reason that should change within the shell wall?
Yes, because the shell is not vacuum. However, after thinking this over, I may have approached this wrong by focusing on the tangential metric components. Let me try a different tack.
First, I want to be clear about how the Schwarzschild "r" coordinate is defined. It is defined such that the area of a sphere at "radius" r is 4 \pi r^{2}. Now consider two spherical "shells" in Schwarzschild spacetime, one at radius r and the other at radius r + dr. These "shells" are not actual physical objects; they are just a way of helping to visualize the physics involved. The area of the inner "shell" is 4 \pi r^{2}, and that of the outer is 4 \pi \left( r + dr \right)^{2}. Suppose we put a ruler between the two "shells" and measure the distance between them. What will it be? If space were "flat", it would simply be dr; but because of the way the Schwarzschild metric works, the actual distance will be
\frac{dr}{1 - \frac{2M}{r}}
(assuming that dr << r, so the metric at r is, to a good enough approximation, the metric at r + dr as well).
So the "anisotropy" you are talking about is "real", in the sense that there is something non-Euclidean about the space between the shells. (However, see my footnote about this at the end of this post.) We can carry this all the way down to the outer surface of the actual shell, the non-vacuum region. Call that outer radius r_{o}. The area of that outer surface is 4 \pi r_{o}^{2}. If we imagine a "shell" (an imaginary one this time) at radius r_{o} + dr, its area would be 4 \pi \left( r_{o} + dr \right)^{2}, but the distance between the two "shells" would be
\frac{dr}{1 - \frac{2M}{r_{o}}}
However: now imagine a spherical "shell" slightly *below* the outer surface of the non-vacuum region, at r_{o} - dr. Its area will be 4 \pi \left( r_{o} - dr \right)^{2} What will the distance between this "shell" and the outer surface of the non-vacuum region be? It will still not be the "Euclidean" distance dr, but something larger; but it will be slightly "less larger" than it would have been if the two shells had been separated by vacuum. If we then continue down through the non-vacuum region, down to its inner radius r_{i}, the distance between "shells" at radius r and radius r + dr, inside the non-vacuum region, will continue to get "less larger" than the "Euclidean" value.
Finally, we reach the inner surface of the non-vacuum region, at r_{i}. The area of that inner surface is 4 \pi r_{i}^{2}, and the area of a "shell" at a slightly larger radius, r_{i} + dr, would be 4 \pi \left( r_{i} + dr \right)^{2}. The distance between these shells, as measured with a ruler, will be just *slightly* larger than dr.
And now, if we consider a "shell" in the vacuum region just inside the inner surface of the non-vacuum region, at radius r_{i} - dr, its area will be 4 \pi \left( r_{i} - dr \right)^{2}, *and* the distance, measured with a ruler, between it and the inner surface of the vacuum region will be exactly dr--no "correction" factor any more. This tells us that the vacuum region inside the inner surface is now "flat"--space there is Euclidean. However, the "potential" there is going to be the same as it is on the inner surface of the shell (because the potential has to be constant throughout the inner vacuum region), and we know that potential is somewhat *less* than that at the outer surface of the vacuum region (as you've already agreed). So the potential in the inner vacuum region is indeed "redshifted" compared to that at infinity. That potential difference no longer shows up in the spatial parts of the metric, but if we compared the rate of time flow in the inner vacuum region to that at infinity, we would find it to be slower, by exactly the same factor as on the inner surface of the non-vacuum region. Another way of saying this is to say that, to put the metric in the interior vacuum region into the standard Minkowski form, we would have to re-scale the time coordinate, compared to that "at infinity", by the "time dilation factor" on the inner surface of the non-vacuum region.
You'll note that I didn't change anything about the tangential metric components at all during any of this; each sphere at "radius" r had the same area as a function of r. However, the metric coefficient g_rr did change, meaning that the relationship between tangential distances and radial distances, expressed as a function of the coordinate r, changed as well, in just the right way to make the metric "flat" in the interior vacuum region.
Footnote: Everything I've said above depends not only on a particular definition of the "r" coordinate, but on a particular definition of simultaneity; basically, what I said above applies in a "surface of constant time" picked out of the global spacetime, and it depends on a particular way of picking out that "surface of constant time", the way that Schwarzschild coordinates pick it out. If we chose a different way of picking surfaces of constant time, we would find different spatial geometries in those surfaces, and the above analysis would proceed differently. For example, if we chose surfaces of constant time the way Painleve coordinates do, the surfaces of constant time in the exterior vacuum region, at least, would be spatially flat--the distance between two "shells" at r and r + dr, where "r" is still defined as the square root of (area of the sphere at "r" divided by 4 pi), would be the "Euclidean" value, dr. That's because the surfaces of constant Painleve time are not the same as the surfaces of constant Schwarzschild time.