# Can a black hole absorb photons whose wavelengths are greater than the BH horizon size?

I am wondering whether a black hole can absorb photons whose wavelengths are greater than the horizon size of the black hole. I have seen different contradicting opinions on the answer. For instance here on PSE, the accepted answer claims that the answer is no, without giving any justification whatsoever. While on physicsforums, I think all the answers are pointing that yes, a black hole should be able to absorb photons whose wavelengths are greater than the size of the horizon of the black hole.

If the answer is yes, then black holes are perfect blackbodies, right? Or is it more complicated, because, like some answer on physicsforums claim, a black hole might have only a small probability to absorb photons whose wavelengths are greater than the horizon size (though no mathematical justification is given), thus even though it might be possible for a BH to absorb such photons, it might still not be a blackbody.

Edit: Here's what I've been told by a QFT specialist:

I think your question is not answerable. "A single photon" requires knowing how to quantize the EM field near a black hole. That is an open problem. And also, "single" particles don't objectively exist in accelerated frames. See the Unruh effect.

I am not satisfied with the current two answers. They do not deal with the above comment at all. Furthermore they seem to contradict each other and they aren't crystal clear.

The photon is an elementary point particle in the standard model of physics. It has energy=$hν$, spin 1 and mass zero. The wavelength, associated with the frequency $ν$, is in the wave function $Ψ$ of the photon, where $Ψ^*Ψ$ is the probability for the photon to be found at a given $(x,y,z,t)$ . It is not the extent of the photon, which is always a point "elementary particle", a quantum mechanical entity.

So if the corresponding wavelength is larger but of order of the crossection area of the black hole horizon, the photon would still have high probability of hitting it. It would have to be a very much larger wavelength to reduce the probability of being captured to small numbers.

Can a black hole absorb photons whose wavelengths are greater than the BH horizon size?

There will always be a probability that a very long wavelength photon will be absorbed by the black hole, given the quantum mechanical definition of a photon. The probability diminishes as the wavelength becomes much larger than the horizon dimension of the black hole.

GR (general relativity) is a classical theory, it describes the dynamics of point-like objects in a curved spacetime. To be applicable to a photon, the wavelength should be negligible if compared to the event horizon dimension of a black hole.

However as a photon proceeds towards the horizon it experiences a blueshifting in the frequency, hence a progressive shortening in the wavelength as measured by stationary observers along the approaching path. That means the photon should be conceived as point-like when crossing the horizon.

As per above arguing the answer is yes, the photon should be absorbed by the black hole whatever the initial wavelength.

Note: Yet, a definite answer may come only from a quantum description of the process, if any.

• The wavelength is much shorter than the photon itself, which includes many-many wavelengths. Soft photons can be absorbed with ordinary piece of matter, so the wavelength is not too representative thing to judge. – Vladimir Kalitvianski May 20 '18 at 16:51
• I'm not sure about "point like as it crosses the horizon". From the view of an external observer yes, but the horizon isn't a solid barrier, and for supermassive BH an observer traversing the horizon may not especially notice they are doing so - nothing specific happens (from other Qs). If that's true for an observer it's also true for a photon. And the frame here is surely not an observer on the external path but the perspective of the photon itself as it crosses the horizon, and the perspective from within the BH/singularity (quite different), so I'm not sure the rationale here can be right. – Stilez Jun 21 '18 at 12:19