PAllen said:
match the Rindler horizon with the BH horizon
Just to belabor this a little more...
The construction using Kruskal coordinates that I gave in post #83 is one way to get to the Rindler viewpoint (just transform the LIF coordinates I obtained to Rindler coordinates). However, there is another way to approach it as well. Let ##\varepsilon = r / R - 1##. We then obtain formulas for the proper acceleration ##a## and the distance above the horizon ##d## in terms of ##\varepsilon##. We have
$$
a = \frac{1}{2 R \left( 1 + \varepsilon \right)^2} \sqrt{\frac{1 + \varepsilon}{\varepsilon}}
$$
For small ##\varepsilon##, we can set ##1 + \varepsilon = 1## to obtain
$$
a = \frac{1}{2 R \sqrt{\varepsilon}}
$$
Now for the distance ##d##; the general formula for that is
$$
d = \int_{R}^{r} \sqrt{\frac{r}{r - R}} dr = \sqrt{r (r - R)} + R \ln \left( \sqrt{\frac{r}{R}} + \sqrt{\frac{r}{R} - 1} \right)
$$
Substituting gives
$$
d = R \left[ \sqrt{\varepsilon (1 + \varepsilon)} + \ln \left( \sqrt{1 + \varepsilon} + \sqrt{\varepsilon} \right) \right]
$$
Making the same small ##\varepsilon## approximation as above, and using the fact that, for small ##x##, ##\ln(1 + x) = x##, we obtain
$$
d = 2 R \sqrt{\varepsilon} = 1 / a
$$
In other words, we have shown that, for stationary observers sufficiently close to the BH horizon, their proper acceleration is the reciprocal of their proper distance above the BH horizon--in other words, the BH horizon is their Rindler horizon.
For reference, the values of ##\varepsilon## for the solution I gave a few posts ago are ##\varepsilon = 10^{-17}## for the rocket, and ##\varepsilon = 2.5 \times 10^{-5}## for buoy #2 at maximum altitude and buoy #3 at minimum altitude. Plugging those values into the above formulas should give the appropriate results.