What happens to the wavelength/frequency of a photon as it passes through an event horizon?

Question

I've asked a similar question about photons and black holes but wanted to rephrase it more specifically, so here goes...

Ever since I learned how a photon's wavelength and frequency are indivisibly linked with the speed of light, I've wondered what this tells us about the speed of light itself. Bring red / blue shifting into the picture and I began to think about the edge cases, like could a photon ever be shifted to infinity (an infinitely long wavelength or infinitely large frequency). One of the answers to my original question mentions that a photon released at exactly the event horizon in a radial direction directly away from the center of the black hole would be stuck there. Using the terminology of limits we would say that 'as the distance from the photon's release point and the event horizon goes to zero', we seem to imply that at the event horizon the photon will not move, this of course violates other laws of physics which is why I'm asking the question. This seems to be a scenario that should illuminate the darkness inside the black hole. I'm not being flippant, I truly believe that there is something about our approaches which we've been missing, for example the study of physical limits like absolute-zero. Is there a relationship between why an atom cannot be cooled to 0K just like a photon can never travel at less than c, through whatever medium it finds itself.

Forgive me and please correct me if I describe this poorly, I don't have the proper vocabulary; my goal is to understand whether a connection exists not write something for a peer-reviewed journal. If atoms cooled to temperatures approaching 0 Kelvin start to form a Bose-Einstein condensate due to the overlapping of their quantum wave functions, would there be a similar effect as photons approach an event horizon (EH)? If photons cannot actually pass the EH, then is the EH more like a physical surface in that since photons cannot pass through the EH are the equations attempting to describe the interior actually nonsensical in the same way it would be to try to divide by zero or find the square root of -1? This leads to other questions like if photons (and I'm assuming atoms too) can not pass the EH then any matter falling into the black hole after the formation of the EH would never reach the singularity and thus the distribution of this new mass would form a sphere near the EH? Would it be correct so say that this new matter 'appears' to add mass to the singularity due to its distribution while never actually being in the singularity? I know this is more than one question but I'm not sure how to break it up into better chunks, suggestions are appreciated.

Answer 1

Answering your questions one by one:

1. Photons are bosons. I do think some phenomena of this sort can form, described by quantum field theory. However I am not aware of any predictions or explanations of the scenario you have mentioned. Maybe I'll try investigating that sometime.
2. Well, in fact light does all the same travel at light speed in the event horizon. However, as they follow their free fall, it just happens that they don't escape from the black hole. The event horizon is no physical surface(e.g. of course photons can fly into a black hole). In fact, locally event horizons are no different from any other point in spacetime. This is a misconception that event horizons or the interiors of black holes are somehow different from other regions of spacetime. In fact, there is no difference. It's just their curvature and thus geodesics, geodesic deviations, are different. Note that in complex and hypercomplex systems you can take the square root of -1.
3. Answering all your other questions here. As I explained, the event horizon is NOT a physical surface. Objects DO fall through the black hole. In fact, event horizons, as with any other surfaces in spacetime of general relativity, are locally the same. So forget the speculation.

Now answering the question in your title: Nothing. Allow me to quote myself... event horizons, as with any other surfaces in spacetime of general relativity, are locally the same. So forget the speculation.

Answer 2

One of the answers to my original question mentions that a photon released at exactly the event horizon in a radial direction directly away from the center of the black hole would be stuck there.

...

we seem to imply that at the event horizon the photon will not move, this of course violates other laws of physics which is why I'm asking the question.

This needs to be addressed.

While it is true that the photon remains at the horizon, you have to come to understand that a photon there is nonetheless moving.

This is best visualized using Kruskal–Szekeres coordinates:

In these coordinates, it is clear that the horizon, the light cone of the event at the origin, is light-like.

And, in these coordinates, it is clear that a photon on the horizon isn't "standing still"; the spatial coordinate does indeed change.

The fact that the Schwarzschild radial coordinate does not change does not imply that a photon there does not move. The Schwarzschild coordinates are singular at the horizon but spacetime is perfectly regular there.

In fact, the Schwarzschild coordinates do not include the horizon except for the event at the origin of the Kruskal diagram, i.e., the Schwarzschild coordinates r=2M, for any value of t, are the coordinates of the origin.

Image credit