PAllen said:
Continuing along a different path than
@PeroK , I've been intermittently following up on the approach described above, using results from the indicated papers. So, using Lemaitre style coordinates, adapted to a "free fall from platform" congruence, discussed at p.8 of the first paper referenced, and changing notation from the paper as follows (all in units where c=G=1, with the idea that mass is expressed in terms of SC radius in light seconds, and spatial units - including SC radial coordinate - are in light seconds, direct computations yield seconds):
- I use P for the platform SC radial coordinate, R for SC radius of the BH
- I express e as used in the paper using a definition given earlier in the paper:
$$P=R/(1-e^2)$$
Then the metric for platform based Lemaitre style coordinates is:
$$ds^2=dT^2-R(1-R/P)^{-1}(1/r-1/P)d\rho^2$$
suppressing angular coordinates since we are treating a purely radial problem. Using results given in the paper for dT and ##d\rho## in terms of SC differentials, one can derive that $$\rho-T=\int (R/r - R/P)^{-1/2} dr \tag {1.1}$$
The f(r) defined by the integral can be evaluated to: $$-\sqrt{\frac {rP} R (P-r)} -\frac {P^{3/2}} {\sqrt R} \arctan(\sqrt{P/r-1}) $$ which can be seen to be equivalent to the formula given in the second paper (eq. 3.10) referenced above (taking ##\rho=0## and expressing as T).
Putting in r=P you verify get zero as desired, and then r=R and r=0 are readily computed as probe proper times for free fall from P to SC radius, and then to singularity.
Note, that in these coordinates, the chosen free fall world line from the platform takes the trivial form ##\rho=0##, T varying. Note that generally, $$\rho-T = costant$$ gives a hovering world line (constant computed as f(r)). In our set up, the constant zero means the hovering platform. This shows the useful fact that proper time along the platform world line matches T coordinate time (just plug r=P into the metric above).
Thus, referring to my diagram, the coordinates of event A (the drop event) are simply ##(\rho,T)=(0,0)##, event of probe reaching horizon are ##(0,-f(R))##, and the limiting coordinate for reaching the singularity are ##(0,-f(0))##.
So far, straight forward. But what I wanted to do next was derive the light path from the platform world line ending on the event of probe at horizon, and also the path from platform world line to probe at singularity. If these were derived, the T coordinates of the start events on the platform world line would directly be proper times for the platform from drop event to corresponding signal times, the second one described being the last signal that can reach the probe.
Unfortunately, here I have hit a major snag. It seems that light paths are very complicated to express in these coordinates, and I am not sure when I will find a way around this. As I can currently express this, I would need to numerically solve a truly complicated differential equation.
I found a way to handle the light paths and can now give a complete alternative solution. Ideally, it should agree with
@PeroK , as to numbers calculated, even if equivalence might be hard to show. But I suspect we might not agree and might not know which is right.
The key to handling light within the framework I started is to combine the platform-Lemaitre T coordinate as used above, with the SC r coordinate, rather than trying to use ##\rho##. We have, from eqn (1.1) above:
$$d\rho-dT=(R/r-R/P)^{-1/2}dr$$ and thus $$d\rho=dT+(R/r-R/P)^{-1/2}dr$$
From the metric given in the quoted post (for light we must have ##ds^2=0##), we get:
$$dT^2=R(1-R/P)^{-1}(1/r-1/P)d\rho^2$$
Substituting and rearranging a bit, gives quadratic:
$$(dr/dT)^2+2\sqrt{R(1/r-1/P)}(dr/dT)+(R/r-1)=0$$
Solving this gives:
$$dr/dT=-\sqrt {R/r-R/P}\pm \sqrt {1-R/P}$$
It turns out that taking the minus sign gives you ingoing light paths, while the plus gives outgoing light. Thus, for r=R, the outgoing light is frozen (derivative is zero). Further, for r<R, both derivatives are negative, implying both light directions are heading for r=0, as expected. Now, what we really want, to combine with earlier results, is T as a function of r, which we'll call g(r). So, we just need to take the reciprocal of the last equation and integrate. This is a surprisingly messy integral (I sought help from an online integrator). After a lot of algebra and simplification, I finally get:
##g(r)=\pm2R \sqrt{1-\frac R P} \ln(\left|\sqrt{1-\frac r P} \mp \sqrt {\frac r R - \frac r P}\right|) ##
##~~~~~~~-\sqrt{RP}(1-\frac {2R} P) \arctan \left( \sqrt{\frac P r -1} \right) \pm r\sqrt {1- \frac R P} ##
##~~~~~~~+\sqrt{rR(1-\frac r P)}+C##
Again, using the lower sign choices gives an ingoing light path, upper signs an outgoing path.
The constant is found by matching the light path to the world line events we previously computed. Thus, for the ingoing light to reach ##(0,-f(0))##, we require ##g(0)=-f(0)##. Then simply ##g(P)## gives the proper time since probe drop to send a light signal that will reach the probe as the probe reaches the singularity. Similarly, to get the platform proper time since probe drop for a signal to reach the probe at the horizon, we pick C to achieve ##g(R)=-f(R)##, and then compute ##g(P)##.
Later, I might compute for realistic numbers for a stellar BH, but for an easily computed case, I take the SC radius, R, as 1 light second, the platform r coordinate as 2 light seconds. Then the formulas will give answers in seconds.
For ##-f(0)##, the probe proper time to reach the singularity, I get simply ##\pi \sqrt 2##. This actually becomes the value of C, and then ##g(P)=\sqrt 2(\pi-1)##.
This is the time the platform must wait, per its own clock, after dropping the probe, to send a signal reaching the probe at its "singular death".
For ##-f(R)##, the probe proper time to reach the horizon, I get ##\sqrt 2 + (\pi/\sqrt 2)##. Then C to get ##g(R)=-f(R)## becomes ##\sqrt 2 + \pi / \sqrt 2 + \ln 2 / \sqrt 2 ##. Finally, we have simply
## g(P) = (\pi + \ln 2)/ \sqrt 2)## as the time the platform should wait (per its own clock) to send a signal to reach probe at horizon crossing.
Thus, you can see how fast everything happens even for a 1 light second BH. You send the signal to catch the probe at the horizon only about 2.7 seconds after dropping the probe, and then wait only another .3 seconds to signal it as it reaches the singularity.