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Let's imagine a pair of particles that is entangled. One (call it $P_1$) is released and then falls to a black hole from a distant $x_0$, (for example $x_0=5r_s$) and velocity $v_0(=1/2c)$, while the other particle ($P_2$) is stationary.

It is more likely for $P_1$ to reach the event horizon (EH) in a finite time ($P_1$ cannot see the end of the universe) (although $P_2$ "see" $P_1$ takes infinite time to reach EH).

Hypothesize that the $P_1$ somehow can change its states when reaching EH, and thus we can measure it by measuring the state of $P_2$ (entangled) without actually "seeing" it enter the black hole. I know it is not how quantum entanglement works, but just let's imagine this is the case.

So, my question is how long is required for $P_1$ to reach EH from the $P_2$'s clock perspective?

P/s: useful link for entering black holes from the falling observer perspective.

Nhat Nguyen
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1 Answers1

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Let's avoid the distraction of bringing quantum entanglement into the argument and rephrase the question in terms of where the falling observer is 'now' according to the outside observer, which I suspect is what you are getting at with your question.

enter image description here

in the above animation, the rotating dark blue line is the Schwrszchild line of simultaneity which is effectively 'now for the outside observer. The light blue line is the world line of a photon. Everything is constrained to remain on the line of simultaneity as the universe evolves. Nothing can get off it. Even the photon has to stay on it. This line of simultaneity arrives at the event horizon after infinite coordinate time. Even if the proper time measured by the infalling observer is just 10 minutes, by the time she arrives at the event horizon, the Sun will have become a red giant, the Andromeda galaxy will have collided with our galaxy, the stars will have burnt up all their fuel and gone out. the universe would have gone dark except maybe for some bright spots due to black holes ending their lives in a last final burst of intense Hawking radiation. The infalling observer does not see any of this, because she only sees her light from her past light cone as can be seen by tracing the photon world line back to its source.

It should be clear from the above that everything takes infinite time to fall to the horizon, regardless of initial height or velocity. A more formal mathematical demonstration can be obtained by analysing the Schwarzschild metric.

If we set $ d \Omega$ to zero we obtain the purely radial form of the metric:

$$dS^2 = (1-r_s/r)c^2 dt^2 - 1/(1-r_s/r) dr^2$$

where $r_s = 2GM/(c^2)$ is the Schwarzschild radius and $r$ is the radial coordinate of the test particle. This metric can be modified to take the location of the external observer ($r_o$) into account:

$$dS^2 = \frac{(1-r_s/r)}{(1-r_s/r_o)}c^2 dt^2 - \frac{(1-r_s/r_o)}{(1-r_s/r)} dr^2$$

For a radially infalling photon, the proper time is zero so we can set $dS=0$ so:

$$0 = \frac{(1-r_s/r)}{(1-r_s/r_o)}c^2 dt^2 - \frac{(1-r_s/r_o)}{(1-r_s/r)}dr^2$$

$$ dt = \frac{1}{c} \left(\frac{(1-r_s/r_o)}{(1-r_s/r)} \right) \ dr$$

For a photon falling from $r_p =r_1$ to $r_p =r_2$ we can carry out the definite integral of the above equation with respect to r using those limits:

$$\Delta t = \frac{(1-r_s/r_o)}{c}\int_{r_2}^{r_1} \left(\frac{1}{(1-r_s/r)} \right) \ dr $$

$$\Delta t= (\Delta r +r_s (\log(r_1-r_s)-\log(r_2-r_s))) \ \times (1-r_s/r_o)/c$$

For $r_2=r_s$ we get $-\log(0) = \infty$ so $\Delta t = \infty$ for any initial location of the ingoing photon outside the event horizon ($r_1 >r_s$) and for any outside observer ($r_o >r_s$).

There is more information from Mathpages for particles with mass falling from a given height or at an initial velocity at infinity here and here.

KDP
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