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The zero point energy of a vacuum is different depending on speed and gravity. The zero point energy of the curved geometry of spacetime near a blackhole is higher compared to a distant observer, so to a distant observer it looks like there’s positive energy radiation where there should be zero energy. That energy difference is a result of the curved spacetime around the black hole which is related to its mass, so therefore the energy comes from the black hole’s mass, and the loss of this positive energy can be crudely thought of as the addition of negative energy.

Samuel Curry
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  • Sort of, except we don't expect to see Hawking radiation unless there's an event horizon. See https://physics.stackexchange.com/a/252236/123208 & https://physics.stackexchange.com/q/564887/123208 – PM 2Ring Jul 11 '20 at 06:53
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    Does this answer your question? An explanation of Hawking Radiation – PM 2Ring Jul 11 '20 at 06:53
  • Is the finite redshift desribed there the result of the intensely curved spacetime? – Samuel Curry Jul 11 '20 at 07:09
  • Sorry, I cant really give a better explanation than what John Rennie wrote. But note that the Hawking radiation is emitted from the space near the event horizon, not exactly at the horizon. And for any light emitted near a black hole (not just Hawking radiation) the closer it starts from the horizon, the more red-shifted it will appear to distant observers. – PM 2Ring Jul 11 '20 at 07:22
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    Hi Samuel. A number of us cooperated to write the answer PM 2Ring refers to. I strongly recommend you read it as it will explain why Hawking radiation occurs in a non-mathematical way. – John Rennie Jul 11 '20 at 07:37

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