As a layperson reading Yes, Stephen Hawking Lied To Us All About How Black Holes Decay, if I understand correctly Hawking radiation is a property of the curvature of space and not specifically of black holes.

So, if any mass produces very tiny amount of Hawking radiation, in the very long term are all objects losing their mass? In the very far future do we expect massless universe?


Hawking radiation is emitted only if an event horizon is present i.e. only if the object is a black hole. So masses like the Earth, or my coffee mug, are not going to evaporate by emitting Hawking radiation.

As you would expect from such a click-bait title Siegel's article is misleading, which is a shame since Siegel is a well respected physicist. Hawking did his calculation using a process called a Bogoliubov transformation. There is a non-mathematical description this in the answers to An explanation of Hawking Radiation though even this will be hard going for the non-physicist. The actual mathematics Hawking used will be utterly impenetrable even for most physicists let alone the general public.

In his original paper Hawking says:

One might picture this negative energy flux in the following way. Just outside the event horizon there will be virtual pairs of particles, one with negative energy and one with positive energy. ...

but then:

It should be emphasized that these pictures of the mechanism responsible for the thermal emission and area decrease are heuristic only and should not be taken too literally.

So he makes it very clear that the explanation using virtual particles is a metaphor not a real process. Siegel's complaint seems to be that Hawking used this metaphor in his popular science book A Brief History of Time, which is true but then if he had gone through the actual maths used it would have been an unpopular science book instead. I am not aware that Hawking used this explanation in any scientific publication.

  • $\begingroup$ Hawking radiation is emitted only if an event horizon is present” - The Hawking radiation requires an apparent horizon, not an event horizon. Similarly the Unruh radiation is predicted in the flat Rindler spacetime where the horizon is apparent while no event horizon exists. In fact, the Hawking radiation prevents the event horizon from forming. An evaporating black hole does not have the event horizon: sciencedirect.com/science/article/pii/S0550321316301274 $\endgroup$ – safesphere Jul 10 '20 at 19:14
  • $\begingroup$ This is a good answer, but one point might be worth clarifying. Quantum radiation ("particles appearing out of the vacuum," poetically speaking) is typically emitted by any collapsing body, whether or not it forms an event horizon. But if an event horizon does form, then part of this radiation continues indefinitely and is independent of the details of the collapse. That's the part we usually mean by Hawking radiation. If a horizon is not formed, then the radiation is only temporary and depends on the details of the collapse, so it doesn't lead to evaporation or the information-loss paradox. $\endgroup$ – Chiral Anomaly Dec 25 '20 at 18:31

The radiation, as a property of the curvature of space is called Unruh effect.

We study the framework of quantum field theory in curved spaces. The Unruh effect and Hawking radiation are described for both Minkowski and anti-de Sitter spaces,and grey body factors are approximated for asymptotically flat black holes.

Hawking radiation is one of the manifestations expected by the curvature of space in General relativity, which is very strong nearing a black hole. The rest is called the Unruh effect, still a research subject.

And Hawking did not lie, afaik and as John describes.

Whether the effect can be seen, as all gravitational interactions are extremely weak, (the energy has to come out from the gravitational field), is highly improbable. The final fate of our present universe is modeled various ways, and one of the models expects everything to dissipate in radiation of various forms.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.