1

Suppose we have a small black hole, maybe $1\ \mathrm{cm}$ in diameter, and a ray of light with a wavelength of $10\ \mathrm{m}$ (or more, if necessary) is exactly aligned with it.

Suppose the black hole is situated at a node of the light's electric field. Is it possible then for the light to simply crest again on the opposite side of the black hole as if it wasn't there?

My thoughts:

  1. Yes black holes massively warp spacetime, but spacetime also gets "unwarped" again far enough from the black hole on the other side. If the two antinodes of the wave would be in unwarped regions of space, can it really see the black hole?

  2. If the node of the wave is at the black hole, then the field is $0$. So what energy then is there to disappear inside the event horizon in the first place?

It intuitively makes sense to me why a light ray with wavelength smaller than a large black hole should disappear into the hole. But I'm not seeing what would happen in this reverse case.

HiddenBabel
  • 1,892

1 Answers1

2

If you are thinking of "a ray" of light which is "aimed directly at" a black hole, then you are neglecting light's wave nature. There is a minimum divergence to any beam of light, and the minimum diameter of any "beam waist" is typically of the same length scale as the wavelength. Your long-wavelength ray is generally wider than your black hole.

With this in mind, think of the effect that a black hole has on an infinite plane wave. In the asymptotically flat region very far from the horizon, light is asymptotically undeflected. If the wavelength is short enough that we can model light as particles, all light with an impact parameter less than $b_c = r_s \cdot 3^{3/2}\,2^{-1}$ is absorbed by the black hole. (Source; see also.) But various scattering trajectories mean that no location downstream of the hole is actually completely in shadow. Your incident plane wave transmits some energy to every point downstream of the black hole, even if the wavelength is short.

If the wavelength is long, as in your question, you have the same sort of phenomenon, but you can think of it more simply. In general, you can't block a wave with an obstacle which is shorter than the wavelength: you get diffraction around the obstacle. Long-wavelength light will have some probability of capture on the event horizon, but you'll have substantial transmission by diffraction around the small absorber.

rob
  • 89,569
  • Do you know, in whatever effective theory is needed to make the prediction, whether probability of single-photon detection at a long distance follows analogously to that for diffraction around obstacles in flat spacetime by propagating the wavefunction across the obstacle? – g s Nov 01 '23 at 15:43
  • I do not. "Diffraction of an electromagnetic plane wave by a black hole" sounds like a great title for a summer undergraduate project, but I don't know the literature. – rob Nov 01 '23 at 16:48