Hawking Radiation: how does a particle ever cross the event horizon?

Question

The heuristic argument for Hawking Radiation is, that a virtual pair-production happens just at the event horizon. One particle goes into the black hole, while the other can be observed as radiation.

I never quite understood this explanation. Say, a virtual pair is created. From the point of view of the "radiation particle" the other particle will need an infinite amount of time in order to reach the event horizon. But the virtual pair production violates energy conservation, so the particles can only exist for a finite amount of time (Heisenberg uncertainty principle). Then, they must annihilate each other again, without emitting any radiation.

I often heard the "explanation" that one particle "tunnels" through the event horizon. But isn't this again a flawed argument? Tunnelling is an effect known from flat spacetime and it makes particles cross potentials that they could not cross classically. How could this help to "jump over" an infinite time interval? For me it seems like superimposing ideas from flat spacetime on curved spacetime without any justification.

While I understand, that the original derivation from Hawking follows a different argument, one should, in principle, be able to attach his results to a physical process. I worked through his derivation and while I must admit that I didn't understand every step, it seems to me that a lot of assumptions have to be made along the way in order to produce the results. Also the whole derivation is based on Quantum Field Theory in curved spacetime, which stands on shaky grounds, since nobody can say, where it is valid and where not - we don't have a complete theory of Quantum Gravity and we simply can't say in which situations QFT in curved spacetime is a good approximation.

As you can see, I am really confused about this issue. Every sort of help is appreciated!

Answer 1

I think your main question is how to reconcile the infinite time-to-reach-horizon needed for a particle created just outside the horizon with the fact that virtual pairs only exist for a short time before re-annihilation?

As you've correctly pointed out, the virtual pair picture is only a heuristic (as Hawking said in his original paper), a more satisfactory explanation is provided by the mapping of vacuum states between free-falling and asymptotically stationary observers. However, perhaps it might help to resolve the heuristic problems if, rather than imagining a highly localized point particle pair being created, one of which travels to the horizon, you instead imagine the fluctuation taking place in the quantum vacuum state, which is a nonlocal entity. So, due to the uncertainties, the vacuum fluctuation includes a field configuration with an outgoing electron plane wave and an infalling positron plane wave (for example). This configuration is picked out as consistent by the geometry and the outgoing particle flux is seen by the asymptotic observer.

Answer 2

Actually I think heuristics are very important and understanding them correctly is critical. I would point point out a Feynman statement that we really don't understand the Pauli Exclusion Principle because we don't have a good heuristic for it. Hell go ask several physicists to describe the proof starting with the Wightman Axioms and they can't do it!

In this case parts of the heuristic are misunderstood.

Examine not a virtual pair created at the event horizon, not near it. First you cannot really say that. Uncertainty in the position means tat is not absolutely clear. But for the moment let us say that a pair is created with it's center of mass at the horizon. What does that say? Well the uncertainty in mass and position of each individual says that sometimes both particles will be formed outside the black. Sometimes both will be formed inside the event horizon. But sometimes one will be formed inside the event horizon, sometimes one will be formed outside. This last circumstance is the one leading to hawking radiation ( with the assumption that a particle formed inside the black hole annihilates with an existing particle inside the black hole).

Answer 3

This is a false paradox in that the particle does not reach the event horizon, regardless of the coordinates used to make the measurements. That is, when measured by any and every coordinate system, a black hole will evaporate from Hawking radiation before a particle can reach the event horizon. For example, as calculated using the Schwarzchild metric, whether measured in coordinate time or local (proper) time, a black hole will evaporate when a particle is at a location outside the event horizon. This suggests it is physically impossible for anything to reach and cross an event horizon. My article provides a more complete explanation: Weller D. "Five fallacies used to link black holes to Einstein’s relativistic space-time." Progress in Physics, 2011, v. 1, 93.