Is there a theorem that we can't see objects getting behind EHs?

[Pony]

Schwarzschild black holes have the property that objects with negligible mass need infinite amount of time to pass through the Event Horizon for any observer outside of the black hole. Thus noone on the outside can be sure that an object has went through the Event Horizon (this works when we talk about lights, or causal structure).

I wonder if this holds in general in classical general relativity, that noone outside of an event horizon can have information of something passing through that event horizon.

My intuition is that it would be weird to watch an object and seeing it to disappear. We getting information of that object would be a (half) closed interval, and not an open one.

Anyone has a counterexample, a solution of GR where an outsider can observe an object falling through an EH?

 

[Ibix]

Schwarzschild black holes have the property that objects with negligible mass need infinite amount of time to pass through the Event Horizon for any observer outside of the black hole.

This isn't really correct - how long it takes depends on your choice of coordinates. The invariant fact is that the horizon crossing never enters the past lightcone of any external observer.

I wonder if this holds in general in classical general relativity, that noone outside of an event horizon can have information of something passing through that event horizon.

That's pretty much the definition of an event horizon - it's the boundary of the region that cannot signal outside. If that isn't the case, it's not an event horizon.

Anyone has a counterexample, a solution of GR where an outsider can observe an object falling through an EH?

I don't think it can be possible. If I drop a pulsing light source towards an event horizon I can, in principle, compute $\tau(\tau')$, the proper time $\tau$ on my clock when a pulse emitted at the proper time $\tau'$ on the source's clock arrives at my location. The smoothness of the manifold guarantees that this will be a smooth function, and we know that a pulse emitted at the horizon will never reach me. Thus the arrival time must smoothly approach infinity as the source approaches the horizon crossing event.

 

[Hill]

This figure from the Gravitation by Misner, Thorne, and Wheeler shows how the falling toward the EH is "seen" in the two frames:

 

[Hill]

it would be weird to watch an object and seeing it to disappear

Actually, the outside observer would see the falling object to disappear, although not as it crosses the EH, but rather as its signal gets red-shifted to a level below a finite sensitivity of the observer's instruments.

 

[Ibix]

Actually, the outside observer would see the falling object to disappear, although not as it crosses the EH, but rather as its signal gets red-shifted to a level below a finite sensitivity of the observer's instruments.

Indeed. Classically, the object is in principle visible forever in the sense that the external observer can always trace a null path back to an event on the object's worldline outside the horizon. But light coming along such lines will be increasingly redshifted and dimmed as the object approaches the horizon, so it fades from view.

 

[PAllen]

Indeed. Classically, the object is in principle visible forever in the sense that the external observer can always trace a null path back to an event on the object's worldline outside the horizon. But light coming along such lines will be increasingly redshifted and dimmed as the object approaches the horizon, so it fades from view.

MTW works an exercise showing how extreme this type of redshift really is. If you take a minimally quantum picture (just temperature of a Planck radiator), then after a 'short' period of time, the expected emission time of the next photon of e.g. 1 km or less wavelength becomes longer than the age of the universe.