TransWikia.com

Could this mean Hawking Radiation is wrong?

Physics Asked by Christine Annette LaBeach on February 4, 2021

Does a virtual particle that enters a black hole also become a real particle just like the one that escaped? If so then the mass taken away from the black hole by the escaping virtual particle would be replaced by the one falling into the black hole so the black hole doesn’t actually lose mass and evaporate away.

3 Answers

There are limits to how good the "virtual pair production" picture of Hawking radiation is. the only critical thing for making the Hawking radiation work is realizing that the vacuum state at infinity is different from vacuum state at the horizon, and because they are different, the horizon looks like blackbody radiation to someone at infinity. There are a large number of ways that that mathematical result can be turned into an intuitive result, but I wouldn't take any of them too literally.

Answered by Jerry Schirmer on February 4, 2021

The black hole has to give energy to create two real particles (by some yet unknown mechanism), and it only gets one of the particles back, the other one escapes, making the black hole lose energy and eventually evaporating.

Answered by Wolphram jonny on February 4, 2021

Does a virtual particle that enters a black hole also become a real particle just like the one that escaped?

Yes. Within this “pair production” picture, this second part of the pair, a particle that falls into the black hole is also real.

If so then the mass taken away from the black hole by the escaping virtual particle would be replaced by the one falling into the black hole so the black hole doesn’t actually lose mass and evaporate away.

That is wrong, because this second real particle has a negative energy (relative to a static asymptotic observer). The energy taken away from the black hole by escaping particle equals up to a sign the energy carried by the second part of the pair. So “absorption” of such negative energy particles is precisely the mechanism for the mass loss of black hole.


Note, that for static (or almost static, such as slowly evaporating non-rotating black hole) spacetimes, particles can have negative energy only inside the black hole horizon, while to escape away from the black hole the first member of the pair must be strictly outside. We thus have a “barrier” separating possible trajectories that must be crossed via quantum tunneling. On the other hand for a rotating black hole there are negative energy trajectories even outside the horizon, within ergosphere. So it is possible to extract rotating energy from a black hole via purely classical processes such as Penrose process.

Answered by A.V.S. on February 4, 2021

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP