Timeline for An explanation of Hawking Radiation
Current License: CC BY-SA 4.0
5 events
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| Oct 22 at 12:27 | comment | added | Níckolas Alves | @James We expect the emission of particles with energy comparable to the black hole's temperature, which is just the usual black body spectrum. Since the energy of a particle is always equal to or larger than its mass, we only expect the emission of massive particles by somewhat hot black holes. For very cold black holes the dominant composition would be of photons, gravitons, etc. | |
| Oct 22 at 2:28 | comment | added | James | Nice answer (+1) Could you elaborate which particles are likely to compose the Hawking radiation of large black holes, please? Such radiation is nearly 0 K, and will evaporate the black hole only when the vacuum temperature drops far below today's 2.7 K, right? At this heat death stage of the universe, it seems unlikely that a $\approx 0 K$ radiation can be composed of abundant W-, W+ or even $e^-, e^+, \gamma$ particles. Then what are the particles that should be radiate-able by the black hole at a Hawking temperature of $\approx 0 K$? | |
| S Jan 15, 2023 at 8:46 | history | suggested | Mark Morales II | CC BY-SA 4.0 |
I fixed some mistakes.
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| Jan 15, 2023 at 8:06 | review | Suggested edits | |||
| S Jan 15, 2023 at 8:46 | |||||
| Mar 22, 2022 at 0:38 | history | answered | Níckolas Alves | CC BY-SA 4.0 |