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A black body absorbs all light/radiation in its reach. According to basic laws of physics, the more energy a body absorbs the more it can emit. Therefore, a black body absorbs all energy directed at it and also emits all energy that's been absorbed.

A black hole is known to absorb all sorts of every (light, radiation..). If it absorbs -all- energy around it, it should in addition emit all energy it has absorbed. But due to its vast gravity noting can "escape", and therefore nothing is actually emitted. Is a black hole still considered a perfect black body?

Disclaimer: I am no physicists, I am a senior year high school student with an interest for physics. I don't know the preferred terminology of things, especially not in English (2nd language), so if anything is unclear I will explain in comments below.

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Yes, black holes are supposedly near-perfect black bodies. They emit thermal radiation called Hawking radiation, which, however, does not originate from beyond the event horizon, but is a consequence of the interaction of the strong gravitational field outside the horizon with the vacuum.

The process is sometimes described as the production of 'virtual' particle pairs due to quantum fluctuations, where one of the particles falls into the black hole, forcing the other one to become 'real'.

A graphic explanation why black holes are only near-perfect black bodies is that they cannot absorb photons with wavelength exceeding the black hole's size.

As a side note, due to the equivalence of gravitational and intertial effects due to acceleration, there's the related Unruh effect for accelerating observers.

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  • $\begingroup$ How can one see that black holes cannot absorb photons whose wavelength is bigger than their (black holes) size? I didn't know black holes had size... $\endgroup$
    – untreated_paramediensis_karnik
    May 14, 2018 at 18:03
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    $\begingroup$ A black hole has a radius that depends on its mass. A plane wave of light incident on the black hole would see the hole as a sort of a circular obstruction and undergo diffraction; the diffraction pattern is the inverse of the circular-aperture patterns that lead to things like the Airy disk. If the light wavelength is short compared to the black hole's size, you can "aim" the light at the hole and not worry about diffraction at the edges. For longer-wavelength light, you cannot aim the light at the hole and keep the plane-wave approximation. $\endgroup$
    – rob ♦
    May 20, 2018 at 15:47
  • $\begingroup$ For a specific example, a one-solar-mass black hole (radius 3km) interacting with 30 kilohertz radiation (wavelength 10km) would produce significant diffraction. $\endgroup$
    – rob ♦
    May 20, 2018 at 15:50

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