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I am familiar with the usual derivation of the fact that it is impossible for someone to observe an object enter a black hole, because of the fact that light gets infinitely redshifted as its source gets closer and closer to the event horizon. But one thing that bothers me is when you think of the problem backwards: if I emit a photon from "infinitely" far away in the direction of the black hole, then when it gets to the event horizon it will be infinitely blueshifted from the perspective of a stationary observer just above the horizon. So it means that the measured energy for the photon will be infinite. My question, than, is whether this simple thought experiment has some implication regarding the geometry around the black hole (since this new energy would be a new source for curvature). I know this might be related to some formal problems regarding black hole scattering and things like that, but I would first like to hear a more conceptual take on this.

Edit: Someone has pointed this question as a possible duplicate of What will the universe look like for anyone falling into a black hole?, but I don't think that addresses exactly the point I am raising here: I am more interested in whether there would be room for a physical consequence caused by the blueshift that light suffers when approaching a black hole, regarding things like curvature etc.

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The error in your premise is that you have an observer who is hovering at the event horizon which is impossible. No observer can hover at the event horizon, once they touch the event horizon they will invariably fall into the black hole and eventually meet the singularity. For such in-falling observers, the light being emitted from far away will not be infinitely blue-shifted as it reaches them.

EDIT: I interpreted the OP question as a question about (coordinate) infinities which occur at the event horizon. It appears from the comments that the question is more about the source of the space-time curvature in GR. In this case, the misconception arises from the use of the "photon energy (in some reference frame)" as the assumed source of gravitational space-time curvature which is not true (as is also stated by Dale in his answer). The photon has well defined 4-momentum (energy-momentum) given by a 4-momentum vector $p^a$. The magnitude of this 4-momentum vector (which is actually 0) will not change as the photon approaches the black hole since this 4-momentum is parallel transported (i.e. it will always have magnitude 0). The energy of the photon is merely the 0-th component of this 4-momentum in some given reference frame. The energy of the photon is frame dependent (just as all energies are) - and so it can not serve as a frame independent description of the source of space-time curvature. The space-time curvature is sourced, instead, by a stress-energy tensor ($T_{ab}$).

Consider an observer in a flat, Minkowski, universe who is moving "infinitesimally close to the speed of light" towards a source of light. That source of light will also be extremely blue-shifted according to this observer. It makes no sense to say that because there are classes of (possible/plausible) observers who move infinitesimally close to the speed of light towards some given light source that "space-time curvature" or "gravity" should care about it.

The class of observers hovering just above the event horizon are like the class of observers moving infinitesimally fast towards a light source in the above analogy. The (impossible) class of observers hovering at the event horizon would be the (also impossible) class of observers moving at the speed of light towards the light source in the above analogy.

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    $\begingroup$ The OP clearly mentioned that the observer is just above, not at, the horizon. $\endgroup$ – Avantgarde Sep 25 '18 at 18:01
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    $\begingroup$ I don't quite think that answers my question completely, because you could still think of a stationary observer arbitrarily close to the event horizon, and even though the blueshift would be technically finite, I would be tempted to believe that I could make it large enough so that it could in principle make a difference to the curvature of spacetime. $\endgroup$ – Bruno De Souza Leão Sep 25 '18 at 18:21
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z Sep 26 '18 at 6:34
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Suppose that you start with a black hole of mass M and infinitely far away from the black hole a spherical pulse of light is directed in towards the horizon. Suppose further that the total energy of that pulse is dM when measured at spatial infinity.

Now, as the pulse falls in it gains energy according to the usual gravitational time dilation factor, so the locally measured energy is greater then dM. As mentioned in the question it goes to infinity as the pulse approaches the horizon.

However, the mass of the black hole increases only by dM, and not by infinity. This can be seen in the details of the Komar mass calculation, where the relevant energy is the energy at infinity. In other words, the mass accounts for the local time dilation and so the black hole is related to the mass/energy at infinity and not the local energy

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    $\begingroup$ Nice answer, +1. Another way to say this is that the putatively very high energy of the infalling photon is a totally frame-dependent thing. Suppose an observer A is infalling from infinity and passes by the stationary observer B, just outside the horizon. A says that B is moving outward at $0.99999c$, and that B's opinion about the photon's high energy is due to B's motion. A would find it surprising if the photon's putatively high energy, according to B, had some big effect on the geometry of spacetime. $\endgroup$ – Ben Crowell Sep 26 '18 at 1:15
  • $\begingroup$ Yes, I definitely agree. Your comment also reminds me to mention that the source of gravitation in GR is the full stress energy tensor, not just energy. Particularly for light the momentum components are not negligible. $\endgroup$ – Dale Sep 26 '18 at 1:28

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