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ProfRob
ProfRob
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The falling observer cannot observe or exchange information with the very distant, future universe. They will continue to receive information from the outside universe up to the point they cross the event horizon and beyond. However, the signals they send back will cease to move outwards once they cross the event horizon. Thus the event horizon is a one-way signal membrane.

The falling observer will not see the universe outside in fast-forward as it were. The falling observer will have a well-defined amount of time to receive information prior to reaching the event horizon. This time is not infinite because, even in Schwarzschild coordinates, the light catches up with the falling observer in a finite time. There is then also a very limited amount of time to receive further information from the outside universe once they have crossed the event horizon. 

As a challenge, using Schwarzschild coordinates (which is not the easiest way to do it), I showed in https://physics.stackexchange.com/a/396157/43351 that if we consider something signalling to a falling observer from a distance $r_0$, where the observer falls past them from infinity at $t=0$, that the coordinate time interval $\Delta t$ that they can wait and still send a signal that intercepts the falling observer before they reach the event horizon is (in units where $c=1$) $$\Delta t < \ln \left(\frac{4r_s}{r_0 - r_s}\right)r_s + \left( \frac{2}{3}\left(\frac{r_0}{r_s}\right)^{3/2} + 2\left(\frac{r_0}{r_s}\right)^{1/2} - \ln \left| \frac{\sqrt{r_0/r_s} + 1}{\sqrt{r_0/r_s} -1}\right| - \frac{5}{3}\right)r_s - r_0$$$$\Delta t < \ln \left(\frac{4r_s}{r_0 - r_s}\right)r_s + \left( \frac{2}{3}\left(\frac{r_0}{r_s}\right)^{3/2} + 2\left(\frac{r_0}{r_s}\right)^{1/2} - \ln \left| \frac{\sqrt{r_0/r_s} + 1}{\sqrt{r_0/r_s} -1}\right| - \frac{5}{3}\right)r_s - r_0\ .$$ [Note that this calculation was independently confirmed by Pulsar using Kruskal-Szekeres coordinates (easier) https://physics.stackexchange.com/a/396829/43351 ]

Then, following the falling object on the other side of the event horizon one can calculate an additional time increment of $0.28r_s$ that can be added to the expression above, which accounts for the fact that the falling body can still receive signals from the outside universe between when it crosses the event horizon and meeting the singularity.

For the case of $r_0 \gg r_s$ then $$\Delta t \simeq \frac{2}{3}r_0^{3/2}r_s^{-1/2} - r_0\ .$$ But this is just the freefall time for an object falling from infinity to go from $r_0$ to the singularity minus the time it would take light to travel through flat spacetime from $r_0$ to 0. So this value of $\Delta t$ is pretty much what you would expect if the falling observer had travelled through flat spacetime. Thus from the point of view of the falling observer almost nothing strange seems to happen when looking back (radially) at the (distant) outside universe and the temporal extent of the events they can witness is almost unaltered by the spacetime curvature unless those events are also quite close to the black hole themselves.

Thus the cut-off Schwarzschild coordinate time-tag is $$t = \ln \left(\frac{4r_s}{r_0 - r_s}\right)r_s + \left( \frac{2}{3}\left(\frac{r_0}{r_s}\right)^{3/2} + 2\left(\frac{r_0}{r_s}\right)^{1/2} - \ln \left| \frac{\sqrt{r_0/r_s} + 1}{\sqrt{r_0/r_s} -1}\right| - \frac{5}{3}\right)r_s - r_0 + 0.280r_s\, ,$$ where the clock starts as the falling observer passes the position $r_0$ (from infinity).

If you wished to convert that into a timestamp measured in the proper time coordinate at a stationary position $r_0$ then you would have to use the usual $(1 - r_s/r_0)^{1/2}$ time-dilation correction.

It would be interesting to do some calculations for light on non-radial trajectories.

The falling observer cannot observe or exchange information with the distant, future universe. They will continue to receive information from the outside universe up to the point they cross the event horizon and beyond. However, the signals they send back will cease to move outwards once they cross the event horizon. Thus the event horizon is a one-way signal membrane.

The falling observer will have a very limited amount of time to receive information from the outside universe once they have crossed the event horizon. As a challenge, using Schwarzschild coordinates (which is not the easiest way to do it), I showed in https://physics.stackexchange.com/a/396157/43351 that if we consider something signalling to a falling observer from a distance $r_0$, where the observer falls past them from infinity at $t=0$, that the coordinate time interval $\Delta t$ that they can wait and still send a signal that intercepts the falling observer before they reach the event horizon is (in units where $c=1$) $$\Delta t < \ln \left(\frac{4r_s}{r_0 - r_s}\right)r_s + \left( \frac{2}{3}\left(\frac{r_0}{r_s}\right)^{3/2} + 2\left(\frac{r_0}{r_s}\right)^{1/2} - \ln \left| \frac{\sqrt{r_0/r_s} + 1}{\sqrt{r_0/r_s} -1}\right| - \frac{5}{3}\right)r_s - r_0$$

Then, following the falling object on the other side of the event horizon one can calculate an additional time increment of $0.28r_s$ that can be added to the expression above, which accounts for the fact that the falling body can still receive signals from the outside universe between when it crosses the event horizon and meeting the singularity.

For the case of $r_0 \gg r_s$ then $$\Delta t \simeq \frac{2}{3}r_0^{3/2}r_s^{-1/2} - r_0\ .$$ But this is just the freefall time for an object falling from infinity to go from $r_0$ to the singularity minus the time it would take light to travel through flat spacetime from $r_0$ to 0. So this value of $\Delta t$ is pretty much what you would expect if the falling observer had travelled through flat spacetime. Thus from the point of view of the falling observer almost nothing strange seems to happen when looking back (radially) at the (distant) outside universe and the temporal extent of the events they can witness is almost unaltered by the spacetime curvature unless those events are also quite close to the black hole themselves.

Thus the cut-off Schwarzschild coordinate time-tag is $$t = \ln \left(\frac{4r_s}{r_0 - r_s}\right)r_s + \left( \frac{2}{3}\left(\frac{r_0}{r_s}\right)^{3/2} + 2\left(\frac{r_0}{r_s}\right)^{1/2} - \ln \left| \frac{\sqrt{r_0/r_s} + 1}{\sqrt{r_0/r_s} -1}\right| - \frac{5}{3}\right)r_s - r_0 + 0.280r_s\, ,$$ where the clock starts as the falling observer passes the position $r_0$ (from infinity).

If you wished to convert that into a timestamp measured in the proper time coordinate at a stationary position $r_0$ then you would have to use the usual $(1 - r_s/r_0)^{1/2}$ time-dilation correction.

It would be interesting to do some calculations for light on non-radial trajectories.

The falling observer cannot observe or exchange information with the very distant future universe. They will continue to receive information from the outside universe up to the point they cross the event horizon and beyond. However, signals they send back will cease to move outwards once they cross the event horizon. Thus the event horizon is a one-way signal membrane.

The falling observer will not see the universe outside in fast-forward as it were. The falling observer will have a well-defined amount of time to receive information prior to reaching the event horizon. This time is not infinite because, even in Schwarzschild coordinates, the light catches up with the falling observer in a finite time. There is then also a very limited amount of time to receive further information from the outside universe once they have crossed the event horizon. 

As a challenge, using Schwarzschild coordinates (which is not the easiest way to do it), I showed in https://physics.stackexchange.com/a/396157/43351 that if we consider something signalling to a falling observer from a distance $r_0$, where the observer falls past them from infinity at $t=0$, that the coordinate time interval $\Delta t$ that they can wait and still send a signal that intercepts the falling observer before they reach the event horizon is (in units where $c=1$) $$\Delta t < \ln \left(\frac{4r_s}{r_0 - r_s}\right)r_s + \left( \frac{2}{3}\left(\frac{r_0}{r_s}\right)^{3/2} + 2\left(\frac{r_0}{r_s}\right)^{1/2} - \ln \left| \frac{\sqrt{r_0/r_s} + 1}{\sqrt{r_0/r_s} -1}\right| - \frac{5}{3}\right)r_s - r_0\ .$$ [Note that this calculation was independently confirmed by Pulsar using Kruskal-Szekeres coordinates (easier) https://physics.stackexchange.com/a/396829/43351 ]

Then, following the falling object on the other side of the event horizon one can calculate an additional time increment of $0.28r_s$ that can be added to the expression above, which accounts for the fact that the falling body can still receive signals from the outside universe between when it crosses the event horizon and meeting the singularity.

For the case of $r_0 \gg r_s$ then $$\Delta t \simeq \frac{2}{3}r_0^{3/2}r_s^{-1/2} - r_0\ .$$ But this is just the freefall time for an object falling from infinity to go from $r_0$ to the singularity minus the time it would take light to travel through flat spacetime from $r_0$ to 0. So this value of $\Delta t$ is pretty much what you would expect if the falling observer had travelled through flat spacetime. Thus from the point of view of the falling observer almost nothing strange seems to happen when looking back (radially) at the (distant) outside universe and the temporal extent of the events they can witness is almost unaltered by the spacetime curvature unless those events are also quite close to the black hole themselves.

Thus the cut-off Schwarzschild coordinate time-tag is $$t = \ln \left(\frac{4r_s}{r_0 - r_s}\right)r_s + \left( \frac{2}{3}\left(\frac{r_0}{r_s}\right)^{3/2} + 2\left(\frac{r_0}{r_s}\right)^{1/2} - \ln \left| \frac{\sqrt{r_0/r_s} + 1}{\sqrt{r_0/r_s} -1}\right| - \frac{5}{3}\right)r_s - r_0 + 0.280r_s\, ,$$ where the clock starts as the falling observer passes the position $r_0$ (from infinity).

If you wished to convert that into a timestamp measured in the proper time coordinate at a stationary position $r_0$ then you would have to use the usual $(1 - r_s/r_0)^{1/2}$ time-dilation correction.

It would be interesting to do some calculations for light on non-radial trajectories.

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ProfRob
ProfRob
  • 136.8k
  • 16
  • 303
  • 487

The falling observer cannot observe or exchange information with the distant, future universe. They will continue to receive information from the outside universe up to the point they cross the event horizon and beyond. However, the signals they send back will cease to move outwards once they cross the event horizon. Thus the event horizon is a one-way signal membrane.

The falling observer will have a very limited amount of time to receive information from the outside universe once they have crossed the event horizon. As a challenge, using Schwarzschild coordinates (which is not the easiest way to do it), I showed in https://physics.stackexchange.com/a/396157/43351 that if we consider something signalling to a falling observer from a distance $r_0$, where the observer falls past them from infinity at $t=0$, that the coordinate time interval $\Delta t$ that they can wait and still send a signal that intercepts the falling observer before they reach the event horizon is (in units where $c=1$) $$\Delta t < \ln \left(\frac{4r_s}{r_0 - r_s}\right)r_s + \left( \frac{2}{3}\left(\frac{r_0}{r_s}\right)^{3/2} + 2\left(\frac{r_0}{r_s}\right)^{1/2} - \ln \left| \frac{\sqrt{r_0/r_s} + 1}{\sqrt{r_0/r_s} -1}\right| - \frac{5}{3}\right)r_s - r_0$$

Then, following the falling object on the other side of the event horizon one can calculate an additional time increment of $0.28r_s$ that can be added to the expression above, which accounts for the fact that the falling body can still receive signals from the outside universe between when it crosses the event horizon and meeting the singularity.

For the case of $r_0 \gg r_s$ then $$\Delta t \simeq \frac{2}{3}r_0^{3/2}r_s^{-1/2} - r_0\ .$$ But this is just the freefall time for an object falling from infinity to go from $r_0$ to the singularity minus the time it would take light to travel through flat spacetime from $r_0$ to 0. So this value of $\Delta t$ is pretty much what you would expect if the falling observer had travelled through flat spacetime. Thus from the point of view of the falling observer almost nothing strange seems to happen when looking back (radially) at the (distant) outside universe and the temporal extent of the events they can witness is almost unaltered by the spacetime curvature unless those events are also quite close to the black hole themselves.

Thus the cut-off Schwarzschild coordinate time-tag is $$t = \ln \left(\frac{4r_s}{r_0 - r_s}\right)r_s + \left( \frac{2}{3}\left(\frac{r_0}{r_s}\right)^{3/2} + 2\left(\frac{r_0}{r_s}\right)^{1/2} - \ln \left| \frac{\sqrt{r_0/r_s} + 1}{\sqrt{r_0/r_s} -1}\right| - \frac{5}{3}\right)r_s - r_0 + 0.280r_s\, ,$$ where the clock starts as the falling observer passes the position $r_0$ (from infinity).

If you wished to convert that into a timestamp measured in the proper time coordinate at a stationary position $r_0$ then you would have to use the usual $(1 - r_s/r_0)^{1/2}$ time-dilation correction.

It would be interesting to do some calculations for light on non-radial trajectories.