bossman27 said:
I guess a more specific question would be, at what point exactly does the astronaut's LIF no longer contain the two probes?
Just to wrap up the "puzzle", I'm going to show the math on this. I'll start with the equation for "escape velocity" at a radius r = 2M ( 1 + 2 \epsilon ), where \epsilon << 1, i.e., just above the horizon. The reason for the factor of 2 in front of \epsilon will be clear in a moment.
The equation for "escape velocity" is (note that I will be freely using the binomial expansion and discarding higher-order terms in \epsilon where necessary):
v_e = \sqrt{\frac{2M}{r}} = \sqrt{\frac{2M}{2M (1 + 2 \epsilon )}} = \left( 1 + 2 \epsilon \right)^{-1/2} = 1 - \epsilon
So the factor of 2 above makes the expression for v_e cleaner.
But what does v_e mean, physically? It is the "escape velocity" at radius r, relative to a "static" observer at radius r. That means the escaping object is moving outward at v_e, relative to the static observer; but the astronaut, who is free-falling inward from rest at infinity, is moving *inward* at v_e. So the relative velocity of the escaping object and the static observer is given by the relativistic velocity addition formula:
v = \frac{v_e + v_e}{1 + v_e^2} = \frac{2 (1 - \epsilon)}{1 + \left( 1 - \epsilon \right)^2} = \frac{2 (1 - \epsilon)}{2 - 2 \epsilon + \epsilon^2} = \frac{1}{1 + \epsilon^2 / [ 2 ( 1 - \epsilon ) ]} = \frac{1}{1 + \epsilon^2 / 2} = 1 - \frac{1}{2} \epsilon^2
Now, an interesting point that I didn't raise before: the "puzzle" can actually be stated even more simply, and sharply, than I did in the OP. Here's how: in the astronaut's local inertial frame, the horizon is moving outward at the speed of light. The probe launched outward at escape velocity just above the horizon is moving outward at *less* than the speed of light. But if it's moving slower than the horizon, how can it possibly escape?
The answer, of course, is found by computing the time it would take for the horizon to catch up with the probe, with respect to the local inertial frame. To do that, we need to first compute how far the probe is from the astronaut at the instant the astronaut crosses the horizon. So we need the time with respect to the LIF between the probe's launch and the astronaut crossing the horizon. This is simple since we have the radial coordinate at which the probe is launched; it's just the standard formula for proper time to fall for a Painleve observer from radius 2M (1 + 2 \epsilon ) to radius 2M:
\tau = \frac{2}{3} [ \sqrt{\frac{r}{2M}} ( r ) - 2M ] = \frac{2}{3} [ \sqrt{1 + 2 \epsilon} ( 2M ) ( 1 + 2 \epsilon ) - 2M ] = \frac{4}{3} M [ ( 1 + \epsilon ) ( 1 + 2 \epsilon ) - 1 ] = \frac{4}{3} M ( 3 \epsilon) = 4 M \epsilon
The distance the probe moves in this time is then just v \tau, but since v differs from 1 only by a term quadratic in \epsilon, then to the order of approximation we are using, the distance is simply D = \tau = 4 M \epsilon (the correction term is cubic in \epsilon).
The time it will take for the horizon to catch up to the probe is then T = D / (1 - v), the distance divided by the "closure speed", the difference of the probe's velocity and the horizon's velocity (which is 1). This gives
T = \frac{\tau}{1 - v} = 4 M \epsilon \frac{2}{\epsilon^2} = \frac{8 M}{\epsilon}
Since the horizon starts at time t = 0 in the LIF, and moves outward at speed 1, this will also be the distance from the LIF's origin at which the horizon would catch the probe, according to the LIF. And since \epsilon << 1, we can see that T >> 8M.
Now we need to compare this to the size of the LIF; i.e., we need to answer the question, at what distance (or time) from the origin of the LIF will tidal gravity become non-negligible? The easiest way to quantify this roughly (which, as we will see, is more than enough) is to look at the components of the Riemann curvature tensor at the origin of the LIF; they are all of order M / r^3 = M / (2M)^3 = 1 / 8 M^2. If we write an expression for the metric in the LIF:
g_{ab} = \eta_{ab} + O(x^2)
then the quadratic correction terms will be proportional to the Riemann tensor components; and since they are quadratic, the "distance" (in space or time) at which the corrections are of order unity is given by the inverse square root of the Riemann tensor components. So the acceptable size of the LIF (in space and time), within which the corrections to the metric due to curvature are negligible, is given by
\delta << \sqrt{8} M
But we saw above that the time for the horizon to catch up to the probe was T >> 8 M; and this is *much* greater (by two factors, so to speak) than the size of the LIF. So tidal gravity/spacetime curvature will become non-negligible long before the horizon would be able to catch the probe.
A final note: what happens once tidal gravity becomes non-negligible? One way to approach this is to ask, what if there were no tidal gravity? What would happen? Well, the horizon would catch up to the probe, meaning it would fall below the horizon (and then meet up with the second probe, which was launched from below the horizon). Tidal gravity basically pulls the horizon down so it can't catch the probe. (More precisely, tidal gravity means the hole pulls the horizon down faster than it pulls the escaping probe down, because of their initial spatial separation.) So the horizon starts out catching up to the probe, but tidal gravity causes it to stop catching up and start receding from the probe.
(A good exercise, btw, is to re-do the above analysis in the LIF of the first probe.)