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The following diagram (which I've adapted from Chapter 7 of "Exploring Black Holes" by Taylor, Wheeler & Bertschinger) shows the radial world-line of a free-falling observer into a Schwarzschild black hole (labelled as "world line of a raindrop") and the null geodesics followed by light emitted by that observer, both radially inwards and radially outwards (labelled as ingoing and outgoing flashes). The diagram uses Gullstrand-Painleve coordinates $r, T$.

I have added a further (red) worldline, which represents the trajectory of another free-falling observer (let us call them observer 2) that crosses the event horizon (at $r/M=2$) at a larger value of $T$. Observer 2 is capable of receiving light signals from the first observer up to point E, at which point, the outwardly directed null geodesic intersects with the world line of observer 2 at point G, at $r=0$.

Gullstrand-Painleve freefall observers

So far so good. But imagine, rather than just the first observer drawn, there were a series of freefalling light beacons that crossed the event horizon prior to observer 2 at a range of $T$ values, stretching into the past.

Questions:

Once observer 2 has crossed the event horizon, is it correct that any light they receive from a beacon that has already crossed the horizon would appear to come from behind them? (Where behind means from the direction of increasing $r$).

Up until they reach the event horizon, the light received by observer 2 from all the beacons would come from the direction of decreasing $r$. So does that mean that as observer 2 crosses the event horizon they essentially "pass through" the images of all the beacons at once? And in fact, must they "pass through" the outward facing images of everything else that has previously fallen into the black hole? What on Earth would that look like?

Finally, what would be the Doppler shifts of the light received by observer 2 in these cases?

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3 Answers 3

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The angle at which a lightlike worldline crosses observer 2's worldline in the diagram should not be interpreted as the direction in which obsever 2 must look to see the light.

Here's one way to understand it. Consider two infalling beacons, one just ahead of observer 2 and one just behind observer 2. At any given point along observer 2's worldline, the diagram will show two lightlike worldlines crossing that point: one from one beacon, and one from the other. Both of these lines will be angled to the left in the $r/M<2$ part of diagram, but one will have a steeper slope than observer 2's worldline, and one will have a shallower slope. The relative slopes tell us which way observer 2 needs to look in order to see that light.

This is consistent with the equivalence principle: in a sufficiently small neighborhood of any given observer, spacetime is essentially flat. This is true at every point in the Gllstrand-Painlevé diagram (except at the curvature singularity, where "sufficiently small" collapses to an empty set). The fact that all causal worldlines are angled to the left in the $r/M<2$ part of the diagram is just an artifact of trying to depict the whole curved spacetime in a flat two-dimensional space. We could draw Minkowski space in a similarly distorted way, if we wanted to. The appearance of the diagram depends on which coordinate system we use, and no matter what coordinate system we use, the diagram cannot faithfully represent the geometry — partly because the geometry is curved, and partly because it has Lorentzian signature.

To answer the first question directly: If a beacon crosses the horizon before observer 2 does, then observer 2 will continue looking forward to see the light from that beacon. If a beacon crosses the horizon after observer 2 does, then observer 2 will continue looking behind to see the light from that beacon. The observer never passes through the image of any of the beacons, not even when the observer hits the singularity. In fact, the observer never sees the leading beacon hit the singularity: the last thing the observer sees is some of the light that the beacon emitted before it hit the singularity. This is consistent with the fact that the singularity isn't a point in space. It's an event (I mean, a locus of events) in the future, and that remains true no matter what coordinate system we use and no matter how we draw the diagram.

The question about Doppler shifts requires some calculation. It depends on how far apart the observer and beacon are. Qualitatively, if the beacon falls in just ahead of the observer, say initially 1 meter away, then the observer won't notice any significant redshift when crossing the horizon (if the black hole is big enough). As they approach the singularity, I expect that the observer would see an increasing amount of redshift due to the differential curvature of spacetime (tidal effects) between the beacon and the observer, but I haven't done the calculation to quantify this.

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    $\begingroup$ To elaborate on your point about geometry: We could figure out which way the infalling observer sees a given light ray to be traveling by finding their orthonormal basis vectors and then taking the inner product of their outward-pointing radial basis vector $(\mathsf{e}_r)^\mu$ with the wave vector $k^\mu$. Any vector such that $k^\mu (\mathsf{e}_r)^\nu g_{\mu \nu}$ < 0 will appear to be ingoing. However, the metric has off-diagonal terms in Gullstrand-Painlevé coordinates, meaning that we can have $k^r < 0$ and still end up with $k^\mu (\mathsf{e}_r)^\nu g_{\mu \nu}$ > 0. $\endgroup$ Jun 30, 2021 at 14:47
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    $\begingroup$ @safesphere there are plenty of other diagrams in Taylor's book that show multiple falling observers. Fig. 3 of Chapter 7 for instance (eftaylor.com/exploringblackholes/… ) - there is no (large) error in what I have sketched. (There maybe slight errors in that the worldlines should not be exactly parallel at the same $T$). $\endgroup$
    – ProfRob
    Jun 30, 2021 at 15:00
  • $\begingroup$ Thanks for this clear explanation of the directionality of the received light. But what about that light emitted at the EH by some beacon (or all of the beacons)? It is only seen by observer 2 as they cross the EH. If observer 2 sees this light approaching them from the forward direction, is it that observer 2 disagrees that it was emitted at the EH by the beacon (couldn't the signal have the $r$ coordinate of emission encoded in it - which can be calculated from the proper time on the beacon's own clock)? $\endgroup$
    – ProfRob
    Jun 30, 2021 at 15:28
  • $\begingroup$ @ProfRob I see, it is plotted in the global coordinates, which are not the coordinates of a falling observer, but a transformed global Schwarzschild coordinates. I’ve deleted my comment. However, in the actual coordinates of the second observer, the first observer does not cross the horizon until the second observer does. So they cross at the same proper time. Thus your diagram does not represent what the second observer actually sees. $\endgroup$
    – safesphere
    Jun 30, 2021 at 15:36
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    $\begingroup$ @ProfRob Of course you know all of that... the point is simply that coordinates that are matched to a spacetime's large-scale properties are not necessarily matched to locally-meaningful properties of the motion of test objects. In general, they can't be, and the situation you're highlighting is a good reminder of this. $\endgroup$ Jul 1, 2021 at 13:00
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As an addendum to Chiral Anomaly's excellent answer, I want to consider the second part of the question:

Up until they reach the event horizon, the light received by observer 2 from all the beacons would come from the direction of decreasing r. So does that mean that as observer 2 crosses the event horizon they essentially "pass through" the images of all the beacons at once? And in fact, must they "pass through" the outward facing images of everything else that has previously fallen into the black hole? What on Earth would that look like?

This actually not any different then the following setup in flat Minkowski space. Suppose our observer sits at the origin, and we have a series of stationary beacons setup along a straight line from the observer. If we follow the past light cone of the observer, it will intersect with all the worldlines of the beacons. So when looking along the line of beacons, you see the superimposed image of all beacons (or more realistically just the first beacon blocking the line of sight to all others). What on Earth would that look like? Well, just look out of the window.

The situation with the series of beacons falling into a black hole is no different. For concreteness sake, lets suppose that the beacons and the observer are all following the same radial trajectory, but shifted by a constant amount along the Killing vector field that is timelike at infinity, i.e. the beacons and the observer have been dropped from the same position at infinity at regular intervals. At any point during the observers one way trip to the singularity, when he looks forward --- i.e. along his past lightcone in the forward direction --- he will see the superimposed images of all beacons. In particular, this is also true when he passes the event horizon (and after, all they way to the singularity). In this regard nothing special happens at the event horizon.

In this setup, we can also easily answer the question about the redshift of the beacons as the observer crosses the event horizon. This is easy because the null ray along the event horizon that connects the observer to all the beacons is actually generated by the Killing vector field used to separate the trajectories. Consequently, the four velocity of the observer parallel propagated (back) along this ray will always be equal to each of the four velocities of the beacons as the ray cross them. Consequently, there will be no observable redshift of any of the images of the beacons that the observers see at the moment he crosses the event horizon.

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  • $\begingroup$ Thanks for this. Suppose the light was encoded with the radial coordinate at which it was emitted (which can be computed by the beacon from its proper time). I now understand that the light received will always appear to come from the "forward" direction. But, the falling observer will see that light labelled as coming from a smaller r coordinate than them when they are outside the EH, the same r coordinate as them at the EH and from a larger r coordinate when inside the EH. Maybe I'm having a brain-freeze but this doesn't seem to be like looking out of the window. $\endgroup$
    – ProfRob
    Jul 1, 2021 at 11:25
  • $\begingroup$ @ProfRob The coordinate $r$ is not a local observable. It being encoded in the light is therefore not very natural. When looking out the window the photons your see are not "encoded with the coordinates" of where they originated. They just have a direction and an energy. $\endgroup$
    – TimRias
    Jul 1, 2021 at 11:44
  • $\begingroup$ In your answer you told me the redshift is zero for photons received from the EH (and is presumably non-zero from elsewhere?). That then encodes where the light was emitted. $r$ is a local measurable outside the EH (you can have a series of "mileposts" that you pass). It is even locally measurable inside the EH if you arrange for a pair of observers to fall in radially but separated by some angle, but in any case can be inferred from the proper time on a carried clock (i.e. a timing signal = a radial coordinate). $\endgroup$
    – ProfRob
    Jul 1, 2021 at 12:11
  • $\begingroup$ @ProfRob In the flat space case, we could have chosen Rindler coordinates such that all the beacons are at Rindler radius r=0 when they cross the observers past lightcone. Based on their internal clock they could also "know" the value of their Rindler radius and encode it in their signal. This only gives a weird result, because we chose weird coordinates. There is nothing intrinsically weird about the local observations. $\endgroup$
    – TimRias
    Jul 1, 2021 at 12:50
  • $\begingroup$ Thanks for your time. +1. I see now that this strangeness is likely a coordinate-dependent thing (isn't it always), but it isn't something I've seen remarked upon elsewhere. $\endgroup$
    – ProfRob
    Jul 1, 2021 at 13:29
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A couple of points not covered in answers by ChiralAnomaly and mmeent:

  • Since before observer crosses the horizon the light from previously fallen beacons would appear redshifted and at the horizon the redshift is zero, after the horizon crossing the light from previously fallen beacons would appear blueshifted (but coming from the same direction) to the observer.

  • At the moment of horizon crossing, the energy flux that reaches the observer from beacons that have fallen long time before ($T_\text{b}-T_\text{o}\gg M$) would be exponentially small. This could be seen as a consequence of continuous exponential stretching of space in radial direction that happens along the null horizon generators, so receiving even a single photon from a beacon fallen long before is an extremely unlikely event.

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  • $\begingroup$ By your last point, are you just saying that because the photons would have to be exactly radial, the specific intensity would just fall away for beacons that crossed further back in time? $\endgroup$
    – ProfRob
    Jul 1, 2021 at 20:00
  • $\begingroup$ @ProfRob: No, I meant it as a statement about wave optics. Or, using your words, photon cannot be exactly radial, because it is also a wave, it necessarily has to extend along the radial dimension. Note, that this exponential stretching is also directly responsible for Hawking radiation. $\endgroup$
    – A.V.S.
    Jul 2, 2021 at 4:25

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