B How Can We Observe Black Holes Growing?

[sha1000]

Again, R is not what you think it is.A will never see the spaceship cross the horizon no. Simply because A cannot see inside the horizon by definition of the horizon. This is also true about the situation in post #39.

As Peter said, just because A cannot see it happen does not mean it doesn’t happen.Yes. The spaceship will cross the horizon (and hit the singularity) in finite proper time.What signals? The signals are sent to A, not to B.

Indeed A is no longer observing the spaceship. All he observers are the signals which are emitted by the spaceship.

And what he can see is that the interval between those signals is increasing and at some point becomes infinitely large, but never completely disappears.

 

[Orodruin]

Indeed A is no longer observing the spaceship. All he observers are the signals which are emitted by the spaceship.

Huh? What would be the difference between these two?

 

[sha1000]

The outside observers never receive a signal from the horizon crossing. This is not the same as the horizon crossing not happening. Many things happen that I do not receive signals from.

So if I understand correctly the observer A will stop to receive the signals. And from this point can he conclude that his friend crossed the horizon?

 

[Dale]

So if I understand correctly the observer A will stop to receive the signals. And from this point can he conclude that his friend crossed the horizon?

He can conclude that his friend crossed the horizon well before that.

 

[Orodruin]

So if I understand correctly the observer A will stop to receive the signals. And from this point can he conclude that his friend crossed the horizon?

No. But he will never receive any signal from the horizon crossing time or later as based on the infalling observer’s clock.

I believe this has now taken a significant detour even if it is still mainly on topic… maybe it should be moved to a separate thread? OP should go back to post #39 and examine it closely.

 

[sha1000]

He can conclude that his friend crossed the horizon well before that.

Thank you. I wanted to know if GR tells us that there is a way for the external observer to conclude when object crosses the horizon. This is the answer I was looking for.

 

[sha1000]

No. But he will never receive any signal from the horizon crossing time or later as based on the infalling observer’s clock.

I believe this has now taken a significant detour even if it is still mainly on topic… maybe it should be moved to a separate thread? OP should go back to post #39 and examine it closely.

Yep. I think I get it better now. Thanks

 

[Orodruin]

Thank you. I wanted to know if GR tells us that there is a way for the external observer to conclude when object crosses the horizon. This is the answer I was looking for.

Careful with your statements. There is no ”when” an object passes the horizon for a distant observer as that would require an unphysical simultaneity convention.

 

[Dale]

Thank you. I wanted to know if GR tells us that there is a way for the external observer to conclude when object crosses the horizon. This is the answer I was looking for.

Careful. “When” something distant happens is a matter of coordinates, not physics.

 

[PeterDonis]

this observer knows that it takes infinite time for an object to reach the horizon (from his point of view).

For an appropriate definition of "time" and "his point of view", yes. But that doesn't mean what you think it means. See below.

He does not need to see it.

This statement is literally correct, but not in the way you think. The outside observer can calculate that the infalling object takes only a finite time by its clock to reach the horizon, and that it continues on inward. So indeed he does not need to see these things happen, to know that they do in fact happen.

The mistake you are making is to think that one particular aspect of the outside observer's "point of view"--the fact that his time coordinate goes to infinity at the horizon--is telling him something about the physics of the infalling object. It's not: it's telling him about the increasing distortion of his "point of view" (more precisely, of Schwarzschild coordinates) as the horizon is approached, culminating in "infinite distortion" at the horizon. It's as if the outside observer concluded, from the fact that a Mercator projection of the Earth's surface assigns an infinite coordinate to the North and South poles, that the North and South poles cannot exist. The problem is in the coordinates.

 

[PAllen]

Careful with your statements. There is no ”when” an object passes the horizon for a distant observer as that would require an unphysical simultaneity convention.

Hmm, convention yes. Unphysical is a matter of taste. I would call simultaneity defined from a family of free falling observers in a manner equivalent to how standard cosmological simultaneity is defined, a physically plausible convention. There is no reason a stationary observer is not allowed to adopt it, and then be able to state, per this convention, exactly when some body crosses the horizon.

 

[Orodruin]

physically plausible convention.

That does not make it physical. It is also not necessary to compute the time on the infalling observer’s clock as it passes the horizon. Furthermore, the Schwarzschild t coordinate is arguably more ”physically plausible” and gives a different answer.

 

[PAllen]

That does not make it physical. It is also not necessary to compute the time on the infalling observer’s clock as it passes the horizon. Furthermore, the Schwarzschild t coordinate is arguably more ”physically plausible” and gives a different answer.

Point remains there is no objective definition of physically plausible. It is a matter of aesthetics. To me, there are many physically plausible conventions possible. And if you want to talk about horizon crossing events in relation to an external stationary observer, you pick one that reaches the horizon.

 

[Orodruin]

Point remains there is no objective definition of physically plausible. It is a matter of aesthetics. To me, there are many physically plausible conventions possible. And if you want to talk about horizon crossing events in relation to an external stationary observer, you pick one that reaches the horizon.

But this is precisely the point. You can obtain any answer you want just by choosing coordinates. This makes the "what is the time for the external observer when the object crosses the horizon" an unphysical one. Sure, you can compute it based on some simultaneity convention, but really, why bother? The same thing goes for cosmology. You can compute the distance to an object "now" (according to standard cosmological choice of FLRW coordinates) but that really holds very little physical significance and it cannot be measured anyway.

 

[PAllen]

But this is precisely the point. You can obtain any answer you want just by choosing coordinates. This makes the "what is the time for the external observer when the object crosses the horizon" an unphysical one. Sure, you can compute it based on some simultaneity convention, but really, why bother? The same thing goes for cosmology. You can compute the distance to an object "now" (according to standard cosmological choice of FLRW coordinates) but that really holds very little physical significance and it cannot be measured anyway.

Of course, there is the invariant transition, for any external observer, of when no part of dropped probe's history is in its causal future anymore. That is, the time after which any light signal sent by the external observer will reach the singularity after the probe has already reached it.

For an 8 solar mass BH, and an external observer hovering at twice the SC radius, this transition is just 242 microseconds after dropping a probe. Any time after this (but not before) it is possible to consider that the probe has reached the singularity.

 

[Dale]

the time after which any light signal sent by the external observer will reach the singularity after the probe has already reached it.

That is a good criterion. Of course, it cannot be measured, just predicted. But with knowledge of the maximum thrust of the vehicle, you could place an upper bound on it.

 

[PAllen]

But this is precisely the point. You can obtain any answer you want just by choosing coordinates. This makes the "what is the time for the external observer when the object crosses the horizon" an unphysical one. Sure, you can compute it based on some simultaneity convention, but really, why bother? The same thing goes for cosmology. You can compute the distance to an object "now" (according to standard cosmological choice of FLRW coordinates) but that really holds very little physical significance and it cannot be measured anyway.

I find I still rebel against the notion that "not invariant" must mean "not physical". IMO, length contraction and time dilation are frame variant physical phenomena.

In the current context, consider yet another example of IMO physically plausible, coordinate dependent ways to state that an object has crossed the horizon at some time for a distant observer:

In GR, while there are no global inertial frames, there are local inertial frames. Then one can extent these to large scale coordinates by geodesically extending the basis (the result is Riemann Normal coordinates defined by a 4-velocity at an event). Then it is true that for such a coordinate system build from a momentarily stationary observers, such an extended coordinate system (though different from standard Schwarzschild coordinates) will never include a horizon crossing event. However, for an observer a light year from a stellar BH, a year after it formed, such coordinates built for an observer moving just 1 meter/second toward the BH, the horizon and singularity will included in the present. Thus, claims that the horizon never forms for distant observers are patently false, unless you somehow want to exclude any observer not completely 'stationary'.

 

[PeterDonis]

I find I still rebel against the notion that "not invariant" must mean "not physical". IMO, length contraction and time dilation are frame variant physical phenomena.

I have to disagree with this. I would say that "length contraction" and "time dilation", if they are to be considered physical phenomena, can and must be defined as invariants. An observer in a particular state of motion can measure invariants that can be appropriately described as the "length contracted length" of an object moving relative to that observer, or the "time dilated clock rate" of a clock moving relative to that observer. Those invariants are unproblematically physical things, and, with proper definition, will in fact correspond to what are usually called the "frame variant" quantities "length contraction" and "time dilation". And with a scenario properly set up along such lines, one can usefully discuss "length contraction" and "time dilation" in terms of invariants.

However, many discussions do not properly put into place or consider the background that needs to be set up to define such concepts in terms of invariants (and many participants in such discussions have no idea either that such a background can be put in place or that it needs to be). And not all coordinate-dependent quantities can even be given a useful correspondence with an invariant.

 

[PeterDonis]

one can extent these to large scale coordinates by geodesically extending the basis (the result is Riemann Normal coordinates defined by a 4-velocity at an event).

I assume you are allowing the metric in terms of such "extended" coordinates to vary from the Minkowski metric? Otherwise this construction will not work as soon as tidal gravity effects become non-negligible.

Also, even allowing the metric to vary, such a coordinate patch can only be extended through a region in which none of the geodesics being used cross. Perhaps this is what you have in mind when you say that no such coordinate patch constructed from the 4-velocity of a static observer will include a horizon crossing event (since radial spacelike geodesics in any such construction would cross at a point on the horizon).

 

[Orodruin]

IMO, length contraction and time dilation are frame variant physical phenomena.

These are concepts that I find confuse students for no real reason as they start obsessing over why the rod does not break in frames other than its rest frame from being pushed together or run in loops around the twin paradox. A bit similar to how we tend to not use relativistic mass.
Are they physical? It depends, as long as there is an operational way to measure them that gives the same result regardless of frame they will be, but they are always accompanied by the caveat of ”as measured by X” or something to that effect. We do not ascribe the measurement of some random observer as the length of an object (although we do define rest length as the length measured in the object’s rest frame).

For the horizon crossing, the same is true, you cannot define the time of horizon crossing from picking a random coordinate system. Only a time in that system, which to my understanding does not really solve the OP’s problem - which partially is wanting to use Schwarzschild t as the time coordinate.

Then one can extent these to large scale coordinates by geodesically extending the basis (the result is Riemann Normal coordinates defined by a 4-velocity at an event).

There generally is no guarantee that such a construction will actually cover the part of spacetime we are interested in. If they did we would not need to worry about using local coordinates at all. You could even end up in a situation where the simultaneities cross, making it possible for events with timelike separation to be time reveresed in the frame and I don’t know what to call that if not unphysical.

 

[PAllen]

I assume you are allowing the metric in terms of such "extended" coordinates to vary from the Minkowski metric? Otherwise this construction will not work as soon as tidal gravity effects become non-negligible.

Of course the metric is not Minkowski. Riemann Normal coordinates are a coordinate family definable in arbitrary pseudo-Riemannian manifold.

Also, even allowing the metric to vary, such a coordinate patch can only be extended through a region in which none of the geodesics being used cross. Perhaps this is what you have in mind when you say that no such coordinate patch constructed from the 4-velocity of a static observer will include a horizon crossing event (since radial spacelike geodesics in any such construction would cross at a point on the horizon).

Fermi-normal coordinates are the ones that often have limited coverage due to geodesic crossing, because one requires spacelike geodesics 4-orthogonal to different events along a world line. Riemann Normal coordinates are defined simply by geodesically extending the tetrad at one event, based on a 4-velocity at that one event. Crossing of geodesics does not arise. There are other reasons coverage may not be global, but not due to geodesic crossing. In the case at hand (and even for a Kerr BH), defining such coordinates for a distant event from an old BH with a 4-velocity slowly moving inward relative to a KVF tangent will include the horizon and singularity in the 'present'.

 

[PAllen]

There generally is no guarantee that such a construction will actually cover the part of spacetime we are interested in. If they did we would not need to worry about using local coordinates at all. You could even end up in a situation where the simultaneities cross, making it possible for events with timelike separation to be time reveresed in the frame and I don’t know what to call that if not unphysical.

You are mixing up Riemann-Normal coordinates with Fermi-Normal coordinates. The problem you describe occurs for the latter. In the case at hand, it occurs for Fermi-Normal coordinates based in a stationary observer precisely because their world line is not inertial.

 

[Orodruin]

Crossing of geodesics does not arise.

This is false. The simplest counter example would be the sphere where all geodesics from a point cross at the opposite point. However, you can also find examples in Schwarzschild spacetime. Take an observer in a circular orbit and an observer in radial outward motion. There will exist a radial geodesic motion exactly such that the observers will meet twice. Once when the radially moving observer is moving away from the black hole and once when moving towards it. Both of these events clearly lie on two geodesics that therefore do cross more than once.

 

[PAllen]

This is false. The simplest counter example would be the sphere where all geodesics from a point cross at the opposite point. However, you can also find examples in Schwarzschild spacetime. Take an observer in a circular orbit and an observer in radial outward motion. There will exist a radial geodesic motion exactly such that the observers will meet twice. Once when the radially moving observer is moving away from the black hole and once when moving towards it. Both of these events clearly lie on two geodesics that therefore do cross more than once.

But those are intersections of two timelike geodesics with different 4-velocities. They wouldn't both be part of the same Riemann-Normal coordinate system. I concede (and stated) there may be limitations on the valid coverage of Riemann-Normal coodinate patch, but not in the case at hand - towards a BH for slowly inwardly moving (relative to KVF) defining 4-velocity.

 

[Orodruin]

They wouldn't both be part of the same Riemann-Normal coordinate system.

Of course they would. The coordinates would be the components of the 4-velocities.

 

[PAllen]

Of course they would. The coordinates would be the components of the 4-velocities.

How? You pick one timelike vector and 3 spacelike vectors, all mutually orthogonal. Each specifies a unique geodesic. Your example is two timelike geodesics intersecting twice. Exotic cases like spacetimes with CTC, or closed spacetimes, can have intersections, but then you still get as much coverage as can be expected. (Even in most spacetimes allowing CTC, they are not geodesics).

 

[Orodruin]

You pick one timelike vector and 3 spacelike vectors, all mutually orthogonal.

No, that would just give you the coordinate axes, but that is not enough to specify a coordinate system. Riemann normal coordinates are based on a map from a subset of the tangent space of one point to the manifold itself, not just four vectors. The four vectors you mention are the basis vectors used for the tangent space and the Riemann normal coordinates defined by a tangent vector are its components in this basis.

 

[PAllen]

No, that would just give you the coordinate axes, but that is not enough to specify a coordinate system. Riemann normal coordinates are based on a map from a subset of the tangent space of one point to the manifold itself, not just four vectors. The four vectors you mention are the basis vectors used for the tangent space and the Riemann normal coordinates defined by a tangent vector are its components in this basis.

Agreed, but that still doesn't explain how the two timelike geodesics you described would be relevant to one Riemann-Normal coordinate system. Each would be part of the definition of a different Riemann-Normal coordinate system.

 

[Orodruin]

Agreed, but that still doesn't explain how the two timelike geodesics you described would be relevant to one Riemann-Normal coordinate system. Each would be part of the definition of a different Riemann-Normal coordinate system.

No. Any tangent vector forms part of the Riemann normal coordinate system (at least until you start dealing with double coverage of points). This is precisely the point of the Riemann normal coordinates. You define the geodesic map from the tangent space of a point p to the manifold itself. Then you introduce coordinates on the tangent space by picking four orthonormal basis vectors of the tangent space. The coordinates of a point in the manifold are then the components in that basis of the tangent vector that maps to that point. This tangent vector is not necessarily (and in most cases not) one of the basis vectors.

The problems with Riemann normal coordinates not being global arise precisely because geodesics do cross.

 

[Orodruin]

To take the sphere as an example: You start at one point $p$ and define the geodesic map $v\to \exp(v)$ from $T_p \mathbb S^2$ to $\mathbb S^2$. The geodesics are the great circles and we can normalise the map such that the antipodal point is reached when $|v| = \pi R$. Taking two unit vectors $X$ and $Y$ in $T_p \mathbb S^2$ defines a map from $T_p \mathbb S^2$ to $\mathbb R^2$ through the projection $v \to (x,y) = (g(v,X),g(v,Y))$, which is basically taking the components of $v$ in terms of $X$ and $Y$.

The map from $T_p \mathbb S^2$ to $\mathbb R^2$ is a bijection and so invertible. To define a point on the sphere you need to specify coordinates $x$ and $y$, corresponding to $v = xX + yY$, which generally is not equal to neither $X$ or $Y$. The point on the sphere to which $v$ maps still needs to be defined by following the geodesic defined by $v$. Where this fails to be a global coordinate system is when all of the tangent vectors of length $\pi R$ map to the same point because the geodesics cross. Even if it is the case on the sphere, it is not necessary for the coordinate axes to cross for the coordinate system to fail at a point where geodesics cross. In the construction above, $v_1 = \pi R(X+Y)/\sqrt 2$ and $v_2 = \pi R X$ correspond to coordinates $(\pi R/\sqrt 2, \pi R/\sqrt 2)$ and $(\pi R, 0)$, respectively. Even if the geodesic of $Y$ did not cross those of $X$, this would be a problem for the coordinate system as two different sets of coordinates map to the same point. It is therefore sufficient that any two geodesics from an event cross at a different event for Riemann normal coordinates to be non-global.

The reverse is however not true. If none of the geodesics cross, then the map from the tangent space to the manifold is invertible. This can still fail to be a global coordinate system (if the map is a injective but not surjective), but geodesics crossing is one of the ways that a coordinate system can fail being global (by being surjective but not injective).

 

[cianfa72]

Even if it is the case on the sphere, it is not necessary for the coordinate axes to cross for the coordinate system to fail at a point where geodesics cross. In the construction above, $v_1 = \pi R(X+Y)/\sqrt 2$ and $v_2 = \pi R X$ correspond to coordinates $(\pi R/\sqrt 2, \pi R/\sqrt 2)$ and $(\pi R, 0)$, respectively. Even if the geodesic of $Y$ did not cross those of $X$, this would be a problem for the coordinate system as two different sets of coordinates map to the same point.

Sorry maybe I didn’t get your point: $v_1$ and $v_2$ given as linear combination of unit vectors $X$ and $Y$ are defined such that $|v_1| =|v_2|= \pi R$. So on the sphere the 'normalized' geodesic map $v\to \exp(v)$ from $T_p \mathbb S^2$ to $\mathbb S^2$ map both to the the same antipodal point w.r.t. point $p$.

Was it just an example on the sphere of two non-coordinate axis that happen to intersect at a common point though ?

 

[Orodruin]

Sorry maybe I didn’t get your point: $v_1$ and $v_2$ given as linear combination of unit vectors $X$ and $Y$ are defined such that $|v_1| =|v_2|= \pi R$. So on the sphere the 'normalized' geodesic map $v\to \exp(v)$ from $T_p \mathbb S^2$ to $\mathbb S^2$ map both to the the same antipodal point w.r.t. point $p$.

Was it just an example on the sphere of two non-coordinate axis that happen to intersect at a common point though ?

It is an example of how the mapping of a vector that is not one of the vectors defining the coordinate system is still highly relevant for what part of the manifold the coordinates cover.

 

[cianfa72]

It is an example of how the mapping of a vector that is not one of the vectors defining the coordinate system is still highly relevant for what part of the manifold the coordinates cover.

Ah ok, so you're basically saying not just the mapping of coordinate vectors $\pi R X$ and $\pi R Y$ may intersect at a common point, actually the same 'issue' extends for the 'exponential mapping' of other vectors elements of $T_p \mathbb S^2$.

 

[PAllen]

No. Any tangent vector forms part of the Riemann normal coordinate system (at least until you start dealing with double coverage of points). This is precisely the point of the Riemann normal coordinates. You define the geodesic map from the tangent space of a point p to the manifold itself. Then you introduce coordinates on the tangent space by picking four orthonormal basis vectors of the tangent space. The coordinates of a point in the manifold are then the components in that basis of the tangent vector that maps to that point. This tangent vector is not necessarily (and in most cases not) one of the basis vectors.

The problems with Riemann normal coordinates not being global arise precisely because geodesics do cross.

Ok, I see my mistake. I was inventing yet another coordinate system as follows:

Construct Riemann Normal coordinates for the SC r-t plane, then cross with S2. The spherical symmetry was “ingrained” in my head, so I didn’t see how I was violating the definition of RN coordinates.

However, my proposed coordinates do have the properties I claimed - they are a global, “natural” extension of a local frame in the case of SC geometry.

Thanks for clarifying the details of RN coordinates.

 

[cianfa72]

The problems with Riemann normal coordinates not being global arise precisely because geodesics do cross.

By the way: is this 'problem' someway related to the 'geodesically complete' property of the manifold endowed with an affine connection ?

 

[PAllen]

By the way: is this 'problem' someway related to the 'geodesically complete' property of the manifold endowed with an affine connection ?

No. The issue raised by @Orodruin would occur, for example, in the spherically symmetric solution for a fluid ball surrounded by vacuum.

 

[Orodruin]

However, my proposed coordinates do have the properties I claimed - they are a global, “natural” extension of a local frame in the case of SC geometry.

Well, my issue still remains that - however natural - it remains a distinct artefact of the coordinates. While you may or may not be able to construct an invariant operational way of measuring this, the statement itself of ”when” horizon crossing occurs remains a coordinate dependent statement.

 

[cianfa72]

No. The issue raised by @Orodruin would occur, for example, in the spherically symmetric solution for a fluid ball surrounded by vacuum.

Ah ok, the geodesically incompleteness might be in cases like the punctured plane for instance.

 

[PAllen]

Well, my issue still remains that - however natural - it remains a distinct artefact of the coordinates. While you may or may not be able to construct an invariant operational way of measuring this, the statement itself of ”when” horizon crossing occurs remains a coordinate dependent statement.

What I want to push back against is that there is any natural sense in which “horizon crossing never occurs per an external observer”, something we get over and over - even some physicists make this absurd claim. Providing physically motivated alternatives to SC time coordinate is one way to do this. And your claim that these are unphysical is both unhelpful and judgmental.

 

[Orodruin]

What I want to push back against is that there is any natural sense in which “horizon crossing never occurs per an external observer”, something we get over and over - even some physicists make this absurd claim. Providing physically motivated alternatives to SC time coordinate is one way to do this. And your claim that these are unphysical is both unhelpful and judgmental.

In Schwarzschild spacetime, the horizon crossing is, by definition, spatially separated or in the causal future of any external observer. This is the invariant physical reality. As I have also made clear, this in no way means that the infalling object freezes on the horizon. The statement that horizon crossing never occurs for an external observer is wrong, not because there is an arbitrariness in how to define your coordinates, but because the arbitrariness itself makes the question ambiguous. I believe that it is the insistence on ascribing physicality to particular coordinates that is unhelpful and biased and the source of many misunderstandings of relativity - both special and general. Relativity ultimately is a theory about spacetime geometry where relevant physical question are invariant (although they may have particular interpretations in particular frames).

 

[Grinkle]

Can one observe expansion of the EH even if one cannot ever get a light signal from whatever fell in confirming that it fell in?

 

[PAllen]

Can one observe expansion of the EH even if one cannot ever get a light signal from whatever fell in confirming that it fell in?

In a sense, yes. There are computer generated images of what you would see against background stars as a neutron star or small BH merges with a bigger one, resulting in a single larger BH visible against background stars. However none of what see comes from at or inside a horizon.

 

[Ibix]

Can one observe expansion of the EH even if one cannot ever get a light signal from whatever fell in confirming that it fell in?

Depends what you mean by "observe". You could measure a black hole's mass by descending to some $r$ coordinate slightly above the horizon and hovering and measuring the acceleration needed to do so. Then dump in a load of mass and repeat. You'll find the same acceleration at a larger $r$. Or you could look at gravitational lensing.

But none of that is really observing the event horizon - it's just measuring the mass. You can't really observe the horizon because it isn't a thing, just a null surface.

 

[Grinkle]

In a sense, yes.

I am very likely missing something deep about curved space-time, but to me, that is the same thing as grossly / coarsely seeing something cross the EH.

The fact that I can't ever get photons from at or beyond the EH and the fact that photons redshift asymptotically as they are emitted from closer and closer to the EH are measurement limitations.

If I can see the EH expand by watching what it obscures or lenses, I observed something cross it with a measurement that doesn't depend on photons from the in-falling object, no? I guess its not really a claim that I saw that "thing" cross the EH - but in principle one can measure the total in-fall over time for a BH and if one has enough information on whatever is falling in one can construct a sequence of the in-falling events.

 

[Grinkle]

@Ibix If I read your post first I wouldn't have bothered with mine - thanks!

 

[Orodruin]

I mean, one issue here is using the Schwarzschild metric and test particles falling into the black hole of that. This is not really a physical scenario because it does not actually describe the spacetime of anything with appreciable mass falling into a black hole. It just happens to be a relatively good approximation for many situations when the contribution of the infalling object to the spacetime geometry is negligible. This is certainly not the situation in the merger of two similar sized black holes.

 

[PAllen]

I am very likely missing something deep about curved space-time, but to me, that is the same thing as grossly / coarsely seeing something cross the EH.

The fact that I can't ever get photons from at or beyond the EH and the fact that photons redshift asymptotically as they are emitted from closer and closer to the EH are measurement limitations.

If I can see the EH expand by watching what it obscures or lenses, I observed something cross it with a measurement that doesn't depend on photons from the in-falling object, no? I guess its not really a claim that I saw that "thing" cross the EH - but in principle one can measure the total in-fall over time for a BH and if one has enough information on whatever is falling in one can construct a sequence of the in-falling events.

Here is a dertailed, accurate, simulation of what would be seen against a stellar background for two equal size BH. It is easy to imagine how it would look for other cases: