In a discussion with my son about absolute zero, we arrived at the conclusion that the event horizon might be the place to look, as it "absorbs?" all energy, including light.

Found this in the community which supports the idea that it's pretty cold to the touch: Lowest temperature possible in the universe?

Don't know enough about blackbody radiation to say, but my understanding of black holes is that anything they can't eat gets stretched, shredded, spun and ejected in jets from the poles, which leads me back to:

If we could get close enough to the event horizon to get a measure free from the influence of the "inedible" stuff's temperature, which I would think would have to be pretty close, just how cold would the "outside" of the event horizon, at the point where it's "absorbing" all matter and energy, including light, be?

Things would be moving, which contradicts absolute zero rules, but if all energy is being absorbed at the EH, we must be pretty close.

Current theory suggests that the "inside surface" of the EH should be violently hot, but the temperature of the "outside surface" would almost certainly have to be very cold, wouldn't it?

  • $\begingroup$ That would be the temperature of Hawking radiation, then, but it's not measurable because the receiver used for the measurement would be far hotter than the black hole itself. $\endgroup$
    – CuriousOne
    Oct 30, 2015 at 6:23

1 Answer 1


With your description, if no radiation comes back from the surface of the black hole, the temperature should be the vacuum classical temperature, 0 Kelvin.

BUT Hawking predicted a radiation coming out from quantum mechanical interactions with the vacuum at the limits of the event horizon.

Hawking showed that quantum effects allow black holes to emit exact black body radiation. The electromagnetic radiation is produced as if emitted by a black body with a temperature inversely proportional to the mass of the black hole.

So it depends on the mass of the black hole what black body radiation spectrum it would radiate, which will define the temperature at the event horizon.

A black hole of one solar mass (M☉) has a temperature of only 60 nanokelvin (60 billionths of a kelvin);

This is still a theoretical prediction as such small temperatures from cosmic bodies cannot be measured.

in fact, such a black hole would absorb far more cosmic microwave background radiation than it emits. A black hole of 4.5 × 10^22 kg (about the mass of the Moon, or about 13 micrometers across) would be in equilibrium at 2.7 kelvin, absorbing as much radiation as it emits. Yet smaller primordial black holes would emit more than they absorb and thereby lose mass.

So only when the cosmic microwave backround radiation cools from the expansion of the universe to such small numbers will larger black holes start diminishing in mass, as is expected, as the radiation leaving from the horizon diminishes the energy of the black hole.

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    $\begingroup$ To make that even more impressive: the total luminosity of a 1 solar mass black hole is 9e-29W... now that would be a fun measurement, indeed. The antenna would have to be cooled to roughly the temperature of the black hole and it would have to cover a substantial part of the entire solid angle... and god forbid even a few atoms should fall into the black hole during the entire measurement. :-) $\endgroup$
    – CuriousOne
    Oct 30, 2015 at 6:38
  • $\begingroup$ @CuriousOne Make sure your antennas can deal with the warped spacetime. $\endgroup$ Nov 1, 2015 at 3:50
  • $\begingroup$ @PyRulez: Yeah! Wouldn't it be fun to fly a 100m-ish antenna structure around a black hole at a fraction of the speed of light? :-) Maybe there is a better way, though, but I can't see it, right now. $\endgroup$
    – CuriousOne
    Nov 1, 2015 at 4:54

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